Q1) 3/7 lies between___ fractions

(a) 4/9,5/9

(b)43/99,4/9

(c)42/99,4/9

(d)41/99,42/99

Q2) if a= bm, then _____ (m>0;a,b>0)

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Q1) 3/7 lies between___ fractions

(a) 4/9,5/9

(b)43/99,4/9

(c)42/99,4/9

(d)41/99,42/99

Q2) if a= bm, then _____ (m>0;a,b>0)

&nb

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A.what is the difference of its 10th and 20th terms?

B.can the differece of any terms of this sequence be 368?Justify

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Good Evening teacher,

In Vedic Maths, adding time, at the last 1.35+3.55=490 and the last 2 digits greater than 60 we added 40 and we got the answer 530. My question how do we get "40" ?

Abhijith C J

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Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

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Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

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അടുത്തടുത്ത രണ്ട് അധി ഒറ്റ സംഖ്യകളുടെ ഗുണനഫലം 399 ആയാൽ സംഖ്യകൾ ഏവ

Gushed bgxbfdgbdkh bx

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Give Examples.....

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Reliable latest lines and angles question 6

#### Question No - 20

In the below figure $AB\left \| CD\left \| EF\: \: and \: \: \: GH\left \| KL$. Find $\angle HKL$ ?

In this question from lines and angles exercise.in the solution it was said that:-

### -----------------------------X----------------------------X------------------------X------------X-----

Now, alternate angles are equal

$\angle CHG=\angle HGN=60^0$

$\angle HGN=\angle KNF=60^0$           [ corresponding angles]

Hence, $\angle KNG=180^{o}-60^{o}=120^{o}$

$\Rightarrow \angle GNK=\angle AKL=120^0$      [ corresponding angles]

$\angle AKH=\angle KHD=25^0$           [ alternative angles]

Therefore, $\angle HKL=\angle AKH+\angle AKL=25+120=145^0$

-------------X-------------------------X--------------------------X------------------------------X-----------------------X----------

In the above said solution where did the 60 degree come from?

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thank you

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#### Question No - 15

Draw these figures, Four equal rhombuses

#### Question No - 16

Draw these figures., Five equal rhombuses:

#### Question No - 17

Draw these figures, Four rhombuses around a square:

#### Question No - 18

Parallelograms on two sides of a square :

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Common difference of an arithmetic sequence is 8 and its one term is 45 can the sum of any 15 terms of this sequence be 2018?

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(b) eqilateral
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## Is zero (0) a rational number ? justify your answer ?

Yes, zero is a rational number

zero can be expressed as    $\frac{0}{5},\frac{0}{26},\frac{0}{100}$  etc

which are in the form of, $\frac{p}{q}$  where p and q are integers and q $\neq$ 0.

## Write the simplest form of a rational number    $\frac{177}{413}$

$\frac{177}{413} = \frac{59\times 3}{59\times 7} = \frac{3}{7}$

## Is   $\frac{\sqrt{98}}{\sqrt{2}}$  a rational number or not ?

$\frac{\sqrt{98}}{\sqrt{2}} =\frac{ \sqrt{49}\times \sqrt{2}}{\sqrt{2}} =\sqrt{49}= 7$

so it is rational number.

## Find the decimal expansion of  $\frac{58}{1000}$

$\frac{58}{1000}$   =   0.058 ( Decimal points is shifted  three places to the left)

## Identify an irrational number among the following numbers: 0.13 ,$0.13\overline{ 15}$ , $0.\overline{ 1315}$ , 0.3013001300013.....

0.13 is a terminating number. So, it is not an irrational number.

$0.13\overline{15}$ = 0.131515......, 15 is repeating continuously so it is not an irrational number.

$0.\overline{ 1315}$ = 0.13151315........is repeating continuously so it is not an irrational number.

0.3013001300013....., non terminating and non recurring decimal. Hence, it is an irrational number. So, 0.3013001300013 is an irrational number.

## Is the product of two irrational number is always an irrational number ?

No, it may be rational or irrational.

## Calculate the decimal which represent the fraction    $\frac{7}{8}$

$\frac{7}{8}=0.875$

## Write the sum of  $2\sqrt{5}$ and $3\sqrt{7}$

sum of $2\sqrt{5}$ and $3\sqrt{7}$ $= 2\sqrt{5}+3\sqrt{7}$

## Insert three rational numbers between $\frac{-1}{\: \: 3}$   and   $\frac{-2}{\: \: 3}$.

$\frac{-1}{\: \: 3} = \frac{-4}{12}$

$\frac{-2}{\: \: 3} = \frac{-8}{12}$

So three rational numbers are

$\frac{-5}{\: \: 3},\frac{-6}{12}\: \: and\: \: \frac{-7}{12}$

## Write a real number which has terminating decimal expansion.

$\frac{31}{125} = 0.248$

## Find two rational numbers between 4 and 5 .

$4 = \frac{4}{5} \times 5\: \: and \: \: 5 = \frac{5}{5}\times 5$

$4 = \frac{20}{5}\: \: and \: \: 5 = \frac{25}{5}$

The numbers are $\frac{21}{5}$ and $\frac{22}{5}$

## Express the rational numbers  $0 .\overline{9}$  in the form  $\frac{p}{q}$ ,  where p and q are integers and $q \neq 0$

Let  X  =  0.999............

10x  =  9.999.......

10x - x  =  (9.999....) – (0.999....)

9x  =  9

X = 1

## Calculate the value of  $2.\overline{9}$  in form of  $\frac{p}{q}$   where  p and q are integers  and   $q \neq 0$

Let,  x   =   $2.\overline{9}$ =   2.9999.....

10x   =   29.999.......

10x –x  =  (29.999....) – (2.999....)

9 x  =  27

$x =\frac{27}{9}$

= 3

## Write the sum of $0.\overline{3}$ and $0.\overline{4}$.

$0.\overline{3}+0.\overline{4}=\left ( 0.333.... \right )+\left ( 0.444.... \right )$

= 0.777...

x  = 0.777

10x  =  7.777

10 x – x  = ( 7.777.... ) – ( 0.777....)

9 x  =  7

$x = \frac{7}{9}$

## Calculate the irrational number 2 and 2.5

Since $\sqrt{5}= 2.236$

hence, the irrationl number between2 and 2.5 is $\sqrt{5}$

## Simplify the number   $\left ( \sqrt{2}+\sqrt{5} \right) ^{2}$

$\left ( \sqrt{2}+\sqrt{5} \right) ^{2}= \left (\sqrt{2} \right )^{2}+\left ( \sqrt{5} \right )^{2}$$+2\times \sqrt{2}\times \sqrt{5}$

$= 2+5+2\sqrt{10}$

$= 7+2\sqrt{10}\: \: \left ( irrational\: number \right )$

## Find the rational numbers between 0.121221222122221...and  0.141441444144441... in the p form, where p and q integers and $q \neq 0$

Two rational numbers between0.121221222122221...and 0.141441444144441...are 0.13 and 0.14

$\frac{13}{10} \: \: and\: \: \frac{14}{100}$

$\frac{13}{100}\: \: and\: \: \frac{7}{50}$

## Find four rational numbers between   $\frac{1}{5}$   and   $\frac{1}{6}$

Since LCM of 5 and 6 is 30

$\frac{1}{6} = \frac{1}{6} \times \frac{5}{5} = \frac{5}{30} \times \frac{5}{5} = \frac{25}{150}$

$\frac{1}{5} = \frac{1}{5} \times \frac{6}{6} = \frac{6}{30} \times \frac{5}{5} = \frac{30}{150}$

Hence , four rational numbers between   $\frac{1}{6}$   and   $\frac{1}{5}$

are $\frac{26}{150} , \frac{27}{150} , \frac{28}{150} , \frac{29}{150}$

## Find six rational numbers between 3 and 4.

let  a = 3 , and b = 4

Here, we find six rational numbers i,e n = 6

$d = \frac{b - a}{n + 1} = \frac{4 - 3}{6 + 1} = \frac{1}{7}$

1st rational number =   $a + d = 3 + \frac{1}{7} = \frac{22}{7}$

2nd rational number =  $a + 2d = 3 + \frac{2}{7} = \frac{23}{7}$

3 rd rational number =  $a + 3d = 3 + \frac{3}{7} = \frac{24}{7}$

4th rational number  =  $a + 4d = 3 + \frac{4}{7} = \frac{25}{7}$

5th rational number  =  $a + 5d = 3 + \frac{5}{7} = \frac{26}{7}$

6th rational number  =   $a + 6d = 3 + \frac{6}{7} = \frac{27}{7}$

So six rational numbers are $\frac{22}{7} , \frac{23}{7} , \frac{24}{7} , \frac{25}{7} , \frac{26}{7} , \frac{27}{7}$

## Insert three rational numbers between  and   $\frac{3}{5}$  and $\frac{5}{7}$

LCM of 5 and 7 is  35

$\frac{3}{5} = \frac{3}{5} \times \frac{7}{7} = \frac{21}{35}$

and     $\frac{5}{7} = \frac{5}{7} \times \frac{5}{5} = \frac{25}{35}$

so      $\frac{21}{35} < \frac{22}{35} < \frac{23}{35} < \frac{24}{35} < \frac{25}{25}$

The required three rational numbers are

$\frac{22}{35} , \frac{23}{35}\: \: and\: \: \frac{24}{35}$

## simlify   $\left ( 5+\sqrt{5} \right )\left ( 5+\sqrt{5} \right )$

$\left ( 5+\sqrt{5} \right )\left ( 5+\sqrt{5} \right )= {5^{2}-\left (\sqrt{5} \right )^{2}}$

$= 25-5$

$= 20$

## $\frac{\sqrt{147}}{\sqrt{75}}$  is not a rational number as $\sqrt{147}$  and $\sqrt{75}$ are not rational. State whether it is true or false. Justify your answer.

false

justification  :     $\frac{\sqrt{147}}{\sqrt{75}} = \sqrt{\frac{147}{75}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$

Which is a rational number

## simplify   $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$   by rationalizing the denominator

$\frac{6-4\sqrt{3}}{6+4\sqrt{3}} \times \frac{6-4\sqrt{3}}{6-4\sqrt{3}}$

$= \frac{(6-4\sqrt{3})^{2})}{36-48}$

$= \frac{36+48-48\sqrt{3}}{-12}$

$= \frac{84-48\sqrt{3}}{-12} =-(7-4\sqrt{3})=4\sqrt{3}-7$

## if $\sqrt{2}$ = 1.414, then, find the value of  $\frac{1}{\sqrt{2 +1}}$

$\frac{1}{\sqrt{2 }+1} \times (\frac{\sqrt{2 }-1}{\sqrt{2}-1})=\sqrt{2}-1$

$=1.414-1$

$= 0.414$

Alternative method

:$\frac{1}{\sqrt{2 }+1} =\frac{1}{\sqrt{2 }+1}\times (\frac{\sqrt{2 }-1}{\sqrt{2}-1})$

$= \frac{\sqrt{2}-1}{\sqrt{2}^{2}-1^{2}}$

$= \frac{\sqrt{2}-1}{2-1} =\frac{1.414-1}{1} = 0.414$

## Taking  $\sqrt{2}=1.414$  and $\pi =3.141$ evaluate  $\frac{1}{\sqrt{2}}+\pi$ upto three places of decimal

$\frac{1}{\sqrt{2}} =\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$= \frac{\sqrt{2}}{2}= \frac{1.414}{2}= 0.707$

$\therefore \frac{1}{\sqrt{2}} +\pi = 0.707 + 3.141 =3.848$

## Find three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$

Since LCM of 7 and 11 is 77

$\frac{5}{7} = \frac{5}{7}$$\times \frac{11}{11} = \frac{55}{77}$

$\frac{9}{11} = \frac{9}{11}\times \frac{7}{7} = \frac{63}{77}$

Hence three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$  are  $\frac{56}{77} , \frac{57}{77} , \frac{58}{77}$

## Rationalize the denominator  $\frac{1}{2\sqrt{7}+3\sqrt{3}}$

$\frac{1}{2\sqrt{7}+3\sqrt{3}} =\frac{1}{(2\sqrt{7}+3\sqrt{3})}\times \frac{(2\sqrt{7}-3\sqrt{3})}{2\sqrt{7}-3\sqrt{3}}$

$=\frac{2\sqrt{7}-3\sqrt{3}}{(2\sqrt{7})^{2}-(3\sqrt{3})^{2}}$

$=\frac{2\sqrt{7}-3\sqrt{3}}{4\times 7-9\times 3}$$=\frac{2\sqrt{7}-3\sqrt{3}}{1}$$= \frac{2\sqrt{7}-3\sqrt{3}}{28-27} = 2\sqrt{7}-3\sqrt{3}$

## Rationalize the denominator of   $\frac{30}{5\sqrt{3}-3\sqrt{5}}$

$\frac{30}{5 \sqrt{3}-3\sqrt{5}} \times \frac{5\sqrt{3 }+3\sqrt{5}}{5\sqrt{3 }+3\sqrt{5}}$ $=\frac{30(5(\sqrt{3 }+3\sqrt{5})}{(5\sqrt{3 })^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3}+3\sqrt{5$

Alternative Method

$\frac{30}{5\sqrt{3}-3\sqrt{5}} = \frac{30}{(5\sqrt{3}-3\sqrt{5})}\times$ $\frac{(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{3}+3\sqrt{5})}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{}3)^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{75-45} =\frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3 }+3\sqrt{5}$

## Simplify   $\frac{1}{1+\sqrt{2}} +\frac{1}{\sqrt{2}+\sqrt{3}} +\frac{2}{\sqrt{3}+\sqrt{5}}$

$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3 }+ \sqrt{5}}$

$=\frac{1}{\sqrt{2}+1}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+ \frac{2}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$=\frac{\sqrt{2}-1}{2-1} +\frac{\sqrt{3}-\sqrt{2}}{3-2} +\frac{2(\sqrt{5}-\sqrt{3})}{5-3}$

$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3} =\sqrt{5}-1$

## Give two rational numbers whose  : (1)  Difference is a rational number (2)  Sum is a rational number (3)  Product is a rational number (4)  Division is a rational  number               Justify also.

Any example and verification of example

$Let\: \: \: m = \frac{4}{5},\: \: \: n = \frac{9}{2}$

$difference\: \: =\frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: \: (Rational number )$

$sum\: \: =\frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: \: (Rational number )$

product          =       4/5 x 9/2 = 36/10 (Rational number)

Division         =       9/2  $\div$ 4/5 = 45/8  ( Rational number)

## Rationalize the denominator of  $\frac{\sqrt{3}+\sqrt{2}}{5+\sqrt{2}}$

$\frac{\sqrt{3} +\sqrt{2}}{5+\sqrt{2} }= \frac{\sqrt{3} + \sqrt{2}}{5+ \sqrt{2}} \times \frac{5-\sqrt{2}}{5 -\sqrt{2}}$

$= \frac{5\sqrt{3} +5\sqrt{2} - \sqrt{6}-2}{25-2}$

$=\frac{5\sqrt{3}+5\sqrt{2} -\sqrt{6}-2}{23}$

## Simplify  $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}} -\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}=\sqrt{18}-\sqrt{12}=3\sqrt{2}-2\sqrt{3}$

$\frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}}=\sqrt{12}-\sqrt{6}=2\sqrt{3}-\sqrt{6}$

$\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{3}}=\sqrt{18}-\sqrt{6}=3\sqrt{2}-\sqrt{6}$

$\therefore$ Given expression  =  $3\sqrt{2}-2\sqrt{3}+2\sqrt{3} -\sqrt{6}-3\sqrt{2}+\sqrt{6}$

= 0

Alternative Method

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6}+\sqrt{3}} -\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$=\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})}+\frac{(3\sqrt{2})}{(\sqrt{6}+\sqrt{3})}$$\times \frac{(\sqrt{6}-\sqrt{3})}{(\sqrt{6}-\sqrt{3)}}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}\times \frac{(\sqrt{6}-\sqrt{2})}{(\sqrt{6}-\sqrt{2})}$

=$\frac{\sqrt{18}-\sqrt{12}}{3-2}+\frac{3\sqrt{12}-3\sqrt{6}}{6-3}-\frac{4\sqrt{18}-4\sqrt{6}}{6-2}$

=$\sqrt{18}-\sqrt{12}+\frac{3(\sqrt{12}-\sqrt{6})}{3}-\frac{4(\sqrt{18}-\sqrt{6})}{4}$

=$\sqrt{18}-\sqrt{12}+\sqrt{12}-\sqrt{6}-\sqrt{18}+\sqrt{6}=0.$

## Simplify  $\frac{5+\sqrt{3}}{7-4\sqrt{3}} \times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}$

$\frac{5+\sqrt{3}}{7-4\sqrt{3}} \times \frac{7+4\sqrt{3}}{7+4\sqrt{3}} =\frac{35+20\sqrt{3}+7\sqrt{3}+12}{49-48}$

$= \frac{47+27\sqrt{3}}{1}$

$and\: \: \: \:\: \: \: \frac{5+\sqrt{3}}{4+4\sqrt{3}} =\frac{(5+\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}$

$=\frac{23-13\sqrt{3}}{1}$

$\frac{5+\sqrt{3}}{7-4\sqrt{3}} -\frac{5+\sqrt{3}}{7+4\sqrt{3}} = (47+27\sqrt{3})-(23-13\sqrt{3})$

$= 24+40 \sqrt{3} = 8(3+5\sqrt{3})$

Alternative Method

$\frac{5+\sqrt{3}}{7-4\sqrt{3}} - \frac{5+\sqrt{3}}{7+4\sqrt{3}}$    $= \frac{(5+\sqrt{3})(7+4\sqrt{3}) -(5+\sqrt{3})(7-4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})}$

$= \frac{35+20\sqrt{3}+7\sqrt{3}+12-35+20\sqrt{3}-7\sqrt{3}+12}{49-48}$

$= \frac{40\sqrt{3}+24}{1} = 8(8+5\sqrt{3}) .$

## Express  the decimal  number  $2.2\overline{18}$  in the form of $\frac{p}{q}$ , where p and q are integers and $q \neq 0$

$let$ $x = 2.2\overline{18} = 2.218181818$....................

10 x  = 22.18181818...........

1000 x = 2218.181818......

1000x – 10x = (2218.181818....) – (22.181818......)

990 x = 2196.00

$x = \frac{2196}{990} =$ $\frac{2\times 3\times 3\times 122}{2\times 3\times 3\times 55} = \frac{122}{55}$

## Express $1.\overline{32}+0.\overline{35}$ in the form $\frac{p}{q}$ , where p and q are integers and $q\neq 0$

$1.3\overline{2}=\frac{119}{90}$

$0.\overline{35}=\frac{35}{99}$

$1.3\overline{2}+0.\overline{35}= \frac{1659}{990}$

## Given two rational numbers whose (1) difference is a rational number (2) sum is a rational number (3) product is a rational number (4) division is a rational number

Any example and verification of example

let m = 4/5, n= 9/2

$\: difference\: = \: \frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: (\: rational \: \: number\: )$

$\: sum\: = \: \frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: (\: rational \: \: number\: )$

$\: product\: = \: \frac{4}{5}\times \frac{9}{2}=\frac{36}{10}\: \: (\: rational \: \: number\: )$

$\: division\: = \: \frac{9}{2}\div \frac{4}{5}=\frac{45}{8}\: \: (\: rational \: \: number\: )$

## show that    $\frac{[x^{a+b}]^{2}\: \: [x^{b+c}]^{2}\: \: [x^{c+a}]^{2}}{(x^{a}x^{b}x^{c})^{4}}= 1$

$\frac{[x^{a+b}]^{2}[x^{b+c}]^{2}[x^{c+a}]^{2}}{(x^{a}x^{b}x^{c})^{4}}$  $= \frac{x^{2a+2b}x^{2b+2c}x^{2c+2a}}{x^{4a}x^{4b}x^{4c}}$

$= \frac{x^{4a+4b+4c}}{x^{4a+4b+4c}}$

$= 1$

## Simplify    $2\: \sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}$

$2\: \sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}-\sqrt[4]{16}$

$= 2\left ( 3^{4} \right )^{\frac{1}{4}}-8\left ( 6^{3} \right )^{\frac{1}{3}}+15\left ( 2^{5} \right )^{\frac{1}{5}}+15-\left ( 2^{4} \right )^{\frac{1}{4}}$

$= 2\times 3-8\times 6+15\times 2+15-2$

$= 6-48+30+15-2$

$= 51-50$

$= 1$

## Find any two irrational numbers between 0.1 and 0.12.

Required to irrational number are :

i)  0.10100100010000......

ii) 0.1020020002000.......

## Simplify  $\frac{\sqrt{2}}{\sqrt{5}+2} -\frac{2}{\sqrt{10}-2\sqrt{2}} +\frac{8}{\sqrt{2}}$

$\frac{\sqrt{2}}{\sqrt{5}+2}-\frac{2}{\sqrt{10}-2\sqrt{2}}+\frac{8}{\sqrt{2}}$

$= \frac{\sqrt{2}}{(\sqrt{5}+2)}\times \frac{({\sqrt{5}-2})}{(\sqrt{5}-2)} -\frac{2}{(\sqrt{10}-2\sqrt{2})}$      $\times \frac{\sqrt{10}+2\sqrt{2}}{\sqrt{10}+2\sqrt{2}}+\frac{8}{\sqrt{2}}\times \frac{\sqrt{2} }{\sqrt{2}}$

$= \frac{\sqrt{10}-2\sqrt{2}}{5-4} -\frac{2 (\sqrt{10}+2\sqrt{2})}{10-8} + \frac{8\sqrt{2}}{2}$

$= \sqrt{10}-2\sqrt{2} - \frac{2(\sqrt{10}+2\sqrt{2})}{2} + 4\sqrt{2}$

$= \sqrt{10}-2\sqrt{2} - {\sqrt{10}-2\sqrt{2}}{} + 4\sqrt{2}$

## Simplify   $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}$

$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}$

$= \frac{1}{\left ( \sqrt{2}+1\right )}\times \frac{\left ( \sqrt{2}-1 \right )}{\left ( \sqrt{2}-1 \right )}$$+\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{3} -\sqrt{2}\right )}$$+\frac{2}{\left ( \sqrt{5} +\sqrt{3}\: \right )}\times \frac{\left ( \sqrt{5}-\sqrt{3} \right )}{\left ( \sqrt{5}-\sqrt{3} \right )}$

$= \frac{\left ( \sqrt{2} -1\right )}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{2\left ( \sqrt{5}-\sqrt{3} \right )}{5-3}$

$= \sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}= \sqrt{5}-1$

## Simplify $\frac{1}{\sqrt{3}+\sqrt{2}} - \frac{2}{\sqrt{5}-\sqrt{3}}- \frac{3}{\sqrt{2}-\sqrt{5}}$

$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}$

$= \frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}$  $\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}$

$= \frac{\sqrt{3}-\sqrt{2}}{1}-\frac{2(\sqrt{5}+\sqrt{3})}{2}-\frac{3(\sqrt{2}+\sqrt{5})}{-3}$

$= \sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5} = 0$

## Simplify   $(\sqrt{3}+1)(1 -\sqrt{12})+\frac{9}{(\sqrt{3}+\sqrt{12})}$

$(\sqrt{3}+1) (1-\sqrt{12})+\frac{9}{(\sqrt{3}+\sqrt{12})}$

$=(\sqrt{3}-6+1-\sqrt{12})+\frac{9}{(\sqrt{12}+\sqrt{3})}\times \frac{(\sqrt{12}-\sqrt{3})}{(\sqrt{12}-\sqrt{3})}$

$=(\sqrt{3}-6+1-\sqrt{12})+ \frac{9(\sqrt{12}+\sqrt{3)}}{12-3}$

$= \sqrt{3}-5-\sqrt{12}+\sqrt{12}-\sqrt{3}$

$= -5.$

## Evaluate   $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$   given that $\sqrt{10}=3.162$

$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} = \frac{(\sqrt{5}+\sqrt{2}) }{\sqrt{5}-\sqrt{2}}\times \frac{(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}$

$= \frac{(\sqrt{5})^{2}+(\sqrt{2})^{2}+2\times \sqrt{5}\times \sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}$

$= \frac{5+2+2\sqrt{10}}{5-2}$

$= \frac{7+2\times 3.162}{3}$

$= \frac{7+6.324}{3} =\frac{13.324}{3}$

$=4.441 (approx)$

## if $a\: = \frac{2^{x-1}}{2^{x-2}}\: \: b= \frac{2^{-x}}{2^{x+1}}\: \: and\: \: a-b= 0,\: \: find\: \: the \: \: value\: \: of \: \: x$

$a-b =0$

$\frac{2^{x-1}}{2^{x-2}}-\frac{2^{-x}}{2^{x+1}}= 0$

$2^{x-1\left ( x-2 \right )}-2^{-x\left ( x+1 \right )}= 0$

$2^{x-1-x+2}-2^{-x-x-1}= 0$

$2^{1}-2^{-2x-1}= 0$

$2^{-2x-1}= 2^{1}$

$-2x-1= 1$

$-2x= 2$

$x= -1$

## $if\: \: \sqrt{2} =1.414 \: \: and \: \: \sqrt{3} = 1.732\: \: then \:\, calculate\: \: \frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$

$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}=\frac{21\sqrt{3}+2\sqrt{2}}{19}$

$=\frac{21(1.732)+2(1.414)}{19}$

$= \frac{39.2}{19}$

$=2.063$

Alternative Method

$\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$     $=\frac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3}-2\sqrt{2})(3\sqrt{3}+2\sqrt{2})}$

$=\frac{12\sqrt{3}+8 \sqrt{2}+9\sqrt{3}-6\sqrt{2}}{27-8}$

$=\frac{21 \sqrt{3}+2\sqrt{2}}{19}$

$=\frac{21\times 1.732+2\times 1.414}{19}$

$=\frac{36.372 +2.828}{19}$

$= \frac{39.2}{19} =2.063$

## If x=  $\frac{1}{3-2\sqrt{2}}$   and y =   $\frac{1}{3+2\sqrt{2}}$   then find the value of x+y+xy.

$x=\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$

$= \frac{3+2\sqrt{2}}{9-8} = 3+2\sqrt{2}$

$y= \frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$=\frac{3+2\sqrt{2}}{9-8} = 3-2\sqrt{2}$

$x+y+xy = 3+2\sqrt{2}+3-2\sqrt{2} + (3+2\sqrt{2}) (3-2\sqrt{2})$

$=6+9-8 =7$

## if  x = 2+$\sqrt{3}$  then find the value of    $x^{2}+\frac{1}{x^{2}}$

$x =2+\sqrt{3}$

$\frac{1}{x} = \frac{1}{2+\sqrt{3}}$

$=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=2-\sqrt{3}$

$\frac{1}{x} =2-\sqrt{3}$

$x +\frac{1}{x} =4$

Squaring both sides we get

$x^{2}+\frac{1}{x^{2}}+2=16$

$x^{2}+\frac{1}{x^{2}}=14$

## If $a =2+\sqrt{5}$   and   $b=\frac{1}{a}$   find  $a^{2}+b^{2}$

$b = \frac{1}{2+\sqrt{5}}$

$=\frac{1}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}}$

$= \frac{2-\sqrt{5}}{-1}=-2+\sqrt{5}$

$a^{2} = (2+\sqrt{5})^{2} =9+4\sqrt{5}$

$b^{2} = (-2+\sqrt{5})^{2} = 9-4 \sqrt{5}$

$a^{2}+b^{2} = 9+4\sqrt{5} +9-4\sqrt{5}$

$=18$

## If $p=5+2\sqrt{6}$   and   $x = \frac{1}{p}$   then what will be the value of   $p^{2}+x^{2}$

$x = \frac{1}{p}= 5-2\sqrt{6 }\: ,\: \: p+x = 5 +2 \sqrt{6} +5-2 \sqrt{6} = 10$

$px = 25-24 =1$

$p^{2} +x^{2} =10^{2}-2\times 1 =98$

Alternative Method

$x = \frac{1}{p} =\frac{1}{5+2\sqrt{6}} =\frac{1}{5+2\sqrt{6}}\times \frac{(5-2\sqrt{6})}{(5-2\sqrt{6})}$

$=\frac{5-2\sqrt{6}}{25-24}= 5-2\sqrt{6}$

$p^{2} +x^{2} = (5+2\sqrt{6})^{2}+(5-2\sqrt{6})^{2}$

$= (5)^{2}+(2\sqrt{6})^{2}+2\times 5\times 2\sqrt{6}+(5)^{2}+(2\sqrt{6})^{2} -2$ $\times 5\times 2\sqrt{6}$

$= 25+24+20\sqrt{6}+25+24-20\sqrt{6} = 98.$

## if  $x =\sqrt{2}-1$ then find the value of $\left (x-\frac{1}{x} \right )^{3}$

$x =\sqrt{2}-1$

$\frac{1}{x} =\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

$\frac{1}{x}= \frac{\sqrt{2}+1}{2-1} =\sqrt{2}+1$

$\therefore \left ( x-\frac{1}{x} \right )^{3} =(\sqrt{2}-1-\sqrt{2}-1)^{3}$

$=(-2)^{3}$

=-8

## Find the value of $\left ( x-\frac{1}{x} \right )^{3} , if x =1+\sqrt{2}$

$x =1+\sqrt{2}$

$\frac{1}{x} =\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{1-2}$

$= \sqrt{2}-1$

$\therefore x-\frac{1}{x} =(1+\sqrt{2})-(\sqrt{2}-1)$

$= 1+ \sqrt{2}-\sqrt{2}+1=2$

$\left ( x-\frac{1}{x} \right )^{3}=2 ^{3}= 8$

Alternative Method

$x = 1+\sqrt{2}$

$\frac{1}{x} =\frac{1}{\sqrt{2 +1}} \times (\frac{\sqrt{2}-1}{\sqrt{2}-1})$

$= (\frac{\sqrt{2}-1}{2-1}) =\sqrt{2}-1$

$\therefore x-\frac{1}{x }= (1+\sqrt{2})-(\sqrt{2}-1)$

$= 1+\sqrt{2} -\sqrt{2}+1 =2$

$\therefore \left ( x - \frac{1}{x} \right )^{3}=2^{3} =8$

## if   $x = 9+4\sqrt{5}$   then find the value of  $\sqrt{x} -\frac{1}{\sqrt{x}}$

$x = 9+4\sqrt{5}$

$= 5+4+4\sqrt{5}$

$=(\sqrt{5}+2)^{2}$

$\sqrt{x} = \sqrt{5}+2$

$\frac{1}{\sqrt{x}} =\sqrt{5}-2$

$\sqrt{x}-\frac{1}{\sqrt{x}} =\sqrt{5}+2-(\sqrt{5}-2)$

$=\sqrt{5}+2-\sqrt{5}+2$

$=4$

Alternative Method

$x = 9+4 \sqrt{5} =5+4+4\sqrt{5}$

$= (\sqrt{5})^{2} + (2)^{2}+2\times 2\sqrt{5}$

$x = (\sqrt{5}+2)^{2}$

$\sqrt{x }=\sqrt{5}+2$

$\frac{1}{\sqrt{x}}=\frac{1}{(\sqrt{5}+2)} \times \frac{(\sqrt{5}-2)}{(\sqrt{5}-2)}$$=\frac{\sqrt{5}-2}{5-4} =\sqrt{5}-2$

$\therefore \sqrt{x }-\frac{1}{\sqrt{x}}=(\sqrt{5}-2)-(\sqrt{5}-2)$
$=\sqrt{5}+2-\sqrt{5}+2 =4$

## Find the values of a and b when   $\frac{5+\sqrt{6}}{5-\sqrt{6}} = a+b \sqrt{6}$

$\frac{5+\sqrt{6}}{5-\sqrt{6}} =\frac{5+\sqrt{6}}{5-\sqrt{6}} \times \frac{5+\sqrt{6}}{5+\sqrt{6}}$

$= \frac{(5+\sqrt{6})^{2}}{(5)^{2}-(\sqrt{6}^{2})}$

$= \frac{25+6+10\sqrt{6}}{25-6}$

$= \frac{31+10\sqrt{6}}{19}$

$\therefore a+b \sqrt{6}=\frac{31}{19} +\frac{10}{19}\sqrt{6}$

Compairing the rational and irrational parts  of both sides we get

$a =\frac{31}{19} \: \: b\: ,=\: \frac{10}{19}$

## Find a and b if   $\frac{1-\sqrt{3}}{1+\sqrt{3}} = a+b$

$a+b =\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$= \frac{1-\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}$

$=\frac{(1-\sqrt{3})^{2}}{1-3}$

$=\frac{1+3-2\sqrt{3}}{-2}$

$=\frac{4-2\sqrt{3}}{-2}$

$a+b = -2+\sqrt{3}$

$a = -2$

$b = \sqrt{3}$

## Find the values of a and b when $a+b\sqrt{15} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\frac{8+2\sqrt{15}}{2}= 4+\sqrt{15}$

$a+b \sqrt{15} =4+\sqrt{15}$

$a= 4 \: ,b = 1$

Alternative method

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{(\sqrt{5}+\sqrt{3})}{\sqrt{5+\sqrt{3}}}$

$= \frac{5+3+2\times \sqrt{5}\times \sqrt{3}}{5-3}$

$=\frac{8+2\sqrt{15}}{2}$

$=\frac{2(4+\sqrt{15})}{2}$

$= 4+\sqrt{15}$

$a+b \sqrt{15} = 4+\sqrt{15}$

Compairing both sides we get

a= 4, b= 1

## if     $\frac{30}{4\sqrt{3}+3\sqrt{2}} = 4\sqrt{3}-a \sqrt{2}$    find the value of a .

$\frac{30}{4\sqrt{3}+3\sqrt{2}} =4\sqrt{3}-a\sqrt{2}$

$\frac{30}{4\sqrt{3}+3\sqrt{2}} \times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}} =4\sqrt{3}-a\sqrt{2}$

$\frac{30(4\sqrt{3}-3\sqrt{2})}{30} = 4\sqrt{3}-a \sqrt{2}$

$4\sqrt{3}-3\sqrt{2}=4\sqrt{3}-a\sqrt{2}$

On compairing     $a= 3$

## Find the value of a and b if   $\frac{\sqrt{2}+1}{\sqrt{2}-1}-\frac{\sqrt{2}-1}{\sqrt{2}+1} = a+\sqrt{2}\: b$

$\frac{\sqrt{2}+1}{\sqrt{2}-1} - \frac{\sqrt{2}-1}{\sqrt{2}+1} = a +\sqrt{2 }\: b$

$\frac{(\sqrt{2}+1)^{2}-(\sqrt{2}-1)^{2}}{(\sqrt{2}-1)(\sqrt{2}+1)} =a+\sqrt{2}\: b$

$\frac{2+1+2\sqrt{2}-2-1+2\sqrt{2}}{2-1} = a+\sqrt{2 }\: b$

$4\sqrt{2}= a+\sqrt{2 b}$

$a= 0, b =4$

## Height of students of  class x are given in the following distribution : Find the model of height  Height (in cm)                  150- 155            155 -160           160- 165            165 - 170            170-175 Number of students              15                      8                     20                     12                      5

Class interval                                Frequency

150 -155                                          15

155 - 160                                           8

160 -165                                          20

165 -170                                         12

170- 175                                          5

Total                                              60

Here,

Modal class = 160 -165

$l = 160, f_{1} = 20, f_{0} = 8, f_{2} = 12, h = 5$

Mode  = $l + \left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$

$= 160 + \left ( \frac{20-8}{40-8-12} \right ) \times 5$

$= 160 \left ( \frac{12}{20} \right )\times 5$

$= 163$

Model height  = 163 cm .

## If     $\frac{3}{4\sqrt{5}-3} +\frac{2}{4\sqrt{5}+\sqrt{3}} = a\sqrt{5 } +b\sqrt{3}$   then find the values of a and b.

$\frac{3}{4\sqrt{5}-\sqrt{3}} + \frac{2}{4\sqrt{5}+\sqrt{3}} = a \sqrt{5}+ b\sqrt{3}$    ............................(1)

$LHS = \frac{3}{4\sqrt{5}-\sqrt{3}}+\frac{2}{4\sqrt{5}+\sqrt{3}}$

$=\frac{3(4\sqrt{2}+\sqrt{3})+2(4\sqrt{5}-\sqrt{3})}{(4\sqrt{5}-\sqrt{3})(4\sqrt{5}+\sqrt{3})}$

$= \frac{12\sqrt{5}+3\sqrt{3}+8\sqrt{5}-2 \sqrt{3}}{80-3}$

$=\frac{20\sqrt{5}+\sqrt{3}}{77} =\frac{20\sqrt{5}}{77}+\frac{\sqrt{3}}{77}$

From eq. (1)

$\frac{20\sqrt{5}}{77}+ \frac{\sqrt{3}}{77} = a\sqrt{5}+b\sqrt{3}$

Compairing on both sides

$a = \frac{20}{77}$  ,  $b = \frac{1}{77}$

## The following frequency distribution shows the number of runs scored by some batsmen of india in one - day cricket mathches: Run scored                      2000-4000           4000-6000           6000-8000         8000-10000           10000-12000 Number of batsmen                9                         8                       10                       2

Here,

Modal class  is  6000- 8000

$f_{0}= 8, f_{1} = 10, f_{2}=2, h =2000$

Mode  = $L + \left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$

$= 6000 + \left ( \frac{10-8}{20-8-2} \right )\times 2000$

$= 6000 + \frac{2}{10}\times 2000$

$= 6000 + 400$

$= 6400$

## If    $x= \frac{\sqrt{p + 2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$   then show that  $qx^{2}-px +q=0$

$x =\frac{\left [ \sqrt{p+2q}+\sqrt{p-2q} \right ]^{2}}{p+2q-p+2q}$

$= \frac{1}{4q} (2p+2\sqrt{p^{2}-4q^{2}})$

$2qx -p =\sqrt{p^{2}-4q^{2}}$

$4q(qx^{2}-px) =4q^{2}$

$qx^{2}-px+q=0$

Alternative Method

$x = \frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}} \times \frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}$

$= \frac{(p+2q)+(p-2q)+2\times \sqrt{p+2q}\times \sqrt{p-2q}}{(p+2q)-(p-2q)}$

$= \frac{2p +2\sqrt{p^{2}-4q^{2}}}{p+2q-p+2q}$

$=\frac{2(p+2\sqrt{p^{2}-4q^{2}})}{4q}$

$2qx = p+\sqrt{p^{2}-4q^{2}}$

$2qx-p=\sqrt{p^{2}-4q^{2}}$

Squaring the both sides we get

$4q^{2}x^{2}+p^{2}-4pqx =p^{2}-4q^{2}$

$4q (qx^{2}-px)=-4q^{2}$

$qx^{2}-px+q =0$

## Rationalize the denominator of $\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{4}}$

$\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{4}}\times \frac{(\sqrt{2}+\sqrt{3})+\sqrt{4}}{(\sqrt{2}+\sqrt{3})+\sqrt{4}}$  $=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{(\sqrt{2}+\sqrt{3})^{2}-{4}} = \frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{(2+3+2\sqrt{6})-4}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{1+2\sqrt{6}}\times \frac{1-2\sqrt{6}}{1-2\sqrt{6}}$

$=\frac{\sqrt{2}+ \sqrt{3}+\sqrt{4}-2\sqrt{12}-2\sqrt{18}-4\sqrt{6}}{1^{2}-(2\sqrt{6})^{2}}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-4\sqrt{3}-6\sqrt{2}-4\sqrt{6}}{1-24}$

$=\frac{-5\sqrt{2}-3\sqrt{3}+2-4\sqrt{6}}{-23}$

$=\frac{+5\sqrt{2}+3\sqrt{3}+4\sqrt{6}-2}{23}$

## Prove that : $\frac{1}{3-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2} =5$

$LHS\: =\: \frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+ \frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$

$=\frac{3+\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}-\frac{\sqrt{8}+\sqrt{7}}{(\sqrt{8})^{2}-(\sqrt{7})^{2}}+\frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$   $-\frac{(\sqrt{6}+\sqrt{5})}{(\sqrt{6})^{2}-(\sqrt{5})^{2}} +\frac{(\sqrt{6}+2)}{(\sqrt{5})^{2}-(2)^{2}}$

$=\frac{3+\sqrt{8}}{9-8}- \frac{(\sqrt{8}+\sqrt{7})}{8-7}+\frac{(\sqrt{7}+\sqrt{6})}{7-6}-\frac{(\sqrt{6}+\sqrt{5})}{6-5}$  $+\frac{(\sqrt{5}+2)}{5-4}\left [ \because a^{2} -b^{2}=(a-b)(a+b) \right ]$

$= 3+\sqrt{8}-\sqrt{8}-\sqrt{7}-\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2$

$= 3+2 =5 =RHS$

## Prove that   $\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}=1.$

$\frac{1}{3+\sqrt{7}}= \frac{1}{3+\sqrt{7}}\times \frac{3-\sqrt{7}}{3-\sqrt{7}} =\frac{3-\sqrt{7}}{9-7} = \frac{3-\sqrt{7}}{2}$

$\frac{1}{\sqrt{7}+\sqrt{5}} = \frac{1}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{7}-\sqrt{5}}{2}$

$\frac{1}{\sqrt{5}+\sqrt{3}} = \frac{1}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{5-3}$

$= \frac{\sqrt{5}-\sqrt{3}}{2}$

$\frac{1}{\sqrt{3}+1} = \frac{1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{3-1}=\frac{\sqrt{3}-1}{2}$

$LHS = \frac{3-\sqrt{7}}{2} + \frac{\sqrt{7}-\sqrt{5}}{2} + \frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}$

$=\frac{3-1}{2}$

$= \frac{2}{2} = 1 = RHS$

## Evaluate   $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$   given that  $\sqrt{5}=2.2 \:\: and\: \sqrt{10}=3.2$

Denominator = $\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5} =3(\sqrt{10}-\sqrt{5})$

$\frac{15}{3(\sqrt{10}\sqrt{5})} = \frac{5\times (\sqrt{10}+\sqrt{5})}{10-5}$

$=\sqrt{10}+\sqrt{5}$

$=3.2+2.2= 5.4$

Alternative Method

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$    $=\frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}$

$=\frac{15}{3\sqrt{10}-3\sqrt{5}}$

$=\frac{15}{3\sqrt{10}-3\sqrt{5}}\times \frac{(\sqrt{10}+\sqrt{5})}{(\sqrt{10}+\sqrt{5})}$

$= \frac{5\times (\sqrt{10}+\sqrt{5})}{10-5}$

$=\frac{5(3.2+2.2)}{5} =5.4$

## Find the  values of a and b in   $\frac{3- \sqrt{5} }{3+2\sqrt{5}} = a\sqrt{5} - \frac{b}{11}$

$a\sqrt{5}-\frac{b}{11} = \frac{(3-\sqrt{5})}{(3+2\sqrt{5})}\times \frac{(3-2\sqrt{5})}{(3-2\sqrt{5})}$

$=\frac{9-6\sqrt{5}-3\sqrt{5}+2\times 5}{(3)^{2}-(2\sqrt{5})^{2}}$

$=\frac{9-9\sqrt{5}+10}{9-20}$

$=\frac{19-9\sqrt{5}}{-11}$

$=\frac{\: \: \: 19}{-11}-\frac{9\sqrt{5}}{-11}=\frac{9\sqrt{5}}{11}-\frac{19}{11}$

$a\sqrt{5}-\frac{b}{11} = \frac{9}{11}\sqrt{5}-\frac{19}{11}$

On Comparing both sides we get

$a = \frac{9}{11}\, \, \; , b =19$

## Find a and b if    $\frac{2\sqrt{5}+\sqrt{3}}{2\sqrt{5}-\sqrt{3}} +\frac{2\sqrt{5}-\sqrt{3}}{2\sqrt{5}+\sqrt{3}}=a +\sqrt{15}\: b$

$LHS=\frac{(2\sqrt{5}+\sqrt{3})^{2}+(2\sqrt{5}-\sqrt{3})^{2}}{(2\sqrt{5}-\sqrt{3})(2\sqrt{5}+\sqrt{3})}$

$=\frac{4\times 5+3+2\times 2\sqrt{5}\times \sqrt{3}+4\times 5+3-2\times 2\sqrt{5}\times \sqrt{3}}{(2\sqrt{5})^{2}-(\sqrt3)^{2}}$

$=\frac{20+3+4\sqrt{15}+20+3-4\sqrt{15}}{20-3}$

$= \frac{46}{17} =\frac{46}{17}+\sqrt{15}\times (0)$

$\therefore \frac{46}{17}+\sqrt{15}(0) =a \sqrt{15}b = RHS$

$Comparing \: \: \: both \: \: \: sides \: \: \: we\: \: \: get$

$a = \frac{46}{17}\: \: , b = 0$

## $if\: x=4-\sqrt{15}\: \: then\: find\: the\: value\: of \left ( x +\frac{1}{x} \right )^{2}$

$x = 4- \sqrt{15}$

$\frac{1}{x} =\frac{1}{4-\sqrt{15}}\times \frac{4+\sqrt{15}}{4+\sqrt{15}}$

$\frac{1}{x} = \frac{4+\sqrt{15}}{16-15}$

$\frac{1}{x} = 4+\sqrt{15}$

$\left ( x+\frac{1}{x} \right )^{2} = (4-\sqrt{15}+4+\sqrt{15})^{2}$

$= (8)^{2}$

$=64$

## If   $a = \frac{2-\sqrt{5}}{2+\sqrt{5}}, b = \frac{2+\sqrt{5}}{2-\sqrt{5}},$   then find  $(a+b)^{3}$

$a = \frac{2-\sqrt{5}}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}} = \frac{(2-\sqrt{5})^{2}}{-1}= -(4-4\sqrt{5}+5)$

$=4\sqrt{5}-9$

$b = \frac{2+\sqrt{5}}{2-\sqrt{5}}\times \frac{2+\sqrt{5}}{2+\sqrt{5}} =\frac{(2+\sqrt{5})^{2}}{-1}$

$=-(4+4\sqrt{5}+5)$

$=-9-4\sqrt{5}$

$a+b = -18$

$(a+b)^{3} =(-18)^{3}= -5832$

Alternative Method

$a+b = \frac{2-\sqrt{5}}{2+\sqrt{5}}+\frac{2+\sqrt{5}}{2-\sqrt{5}}$

$=\frac{(2-\sqrt{5})^{2}+(2+\sqrt{5})^{2}}{(2+\sqrt{5})(2-\sqrt{5})}$

$=\frac{4+5-4\sqrt{5}+4+5+4\sqrt{5}}{4-5}$

$=\frac{18}{-1} = -18$

$(a+b)^{3} =(-18)^{3}= -5832$

## $if\: \: x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3-\sqrt{2}}} \: \: and \: \: y = \frac{\sqrt{3}-\sqrt{2}}{{\sqrt{3}+\sqrt{2}}} \: \: find\: \: the\: \: x^{2 + y^{2}}$

$x =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$and\: \: y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$x^{2} +y^{2} = \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )^{2}+\left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )^{2}$

$= \left ( \frac{3+2+2\sqrt{6}}{3+2-2\sqrt{6}} \right )+\left ( \frac{3+2-2\sqrt{6}}{3+2+2\sqrt{6}} \right )$

$= \frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\frac{5-2\sqrt{6}}{5+2\sqrt{6}}$

$=\frac{(5+2\sqrt{6})^{2}+(5-2\sqrt{6})^{2}}{(5)^{2}-(2\sqrt{6})^{2}}$

$= \frac{25+24+20\sqrt{6}+25+24-20\sqrt{6}}{25-24}$

$= 98$

## If    $x =\frac{\sqrt{3}+1}{\sqrt{3}-1}, y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$     then find the value of   $x^{2}+y^{2}+xy.$

$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}= \frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}= \frac{4+2\sqrt{3}}{2}$

$= 2+ \sqrt{3}$

$y= \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}= \frac{4-2\sqrt{3}}{2}$

$= 2-\sqrt{3}$

$xy =\frac{ \sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}-1}{\sqrt{3}+1} =1$

$\therefore x^{2}+y^{2}+xy = (2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}+1=15$

## if   $x= \frac{\sqrt{2}-1}{\sqrt{2}+1}\: \: and \: \: y= \frac{\sqrt{2}+1}{\sqrt{2}-1}$    then find the value of   $x^{2}+5xy+y^{2}$

$x^{2}+5xy+y^{2}$    $= (x+y)^{2}+3xy$

$= \left [ \frac{\sqrt{2}-1}{\sqrt{2}+1}+\frac{\sqrt{2}+1}{\sqrt{2}-1} \right ]^{2}+$$3\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}+1}{\sqrt{2}-1}$

$=\left [ \frac{2+1-2\sqrt{2}+2+1+2\sqrt{2}}{2-1} \right ]^{2}+3$

$= (6)^{2}+3 =36+3 = 39$

## if      $x=\frac{1}{2-\sqrt{3}}$    then find the value of   $2x^{3 }-2x^{2}+ 7x+5.$

$x =\frac{1}{2-\sqrt{3}}$

$x =\frac{1}{2-\sqrt{3}}\times \frac{(2+\sqrt{3})}{(2+\sqrt{3})}$

$x=\frac{2+\sqrt{3}}{4-3} =2+\sqrt{3}$

$x =2+\sqrt{3}$

$(x+-2)=\sqrt{3}$

$(x-2)^{2} = (\sqrt{3})^{2} =3$

$x^{2}-4x+4=3$

$x^{2}-4x+4-3=0$

$x^{2}-4x+1=0$

$x^{3}-2x^{2}-7x+5$

$x(x^{2}-4x+1)+2(x^{2}-4x+1)+3 =x\times 0+2\times 0+3 =3$

## Prove that  $\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}=1.$

$\frac{1}{\sqrt{4}+\sqrt{5}}=\frac{1}{(\sqrt{5}+\sqrt{4})}\times \frac{(\sqrt{5}-\sqrt{4})}{(\sqrt{5}-\sqrt{4)}}$

$=\frac{\sqrt{5}-\sqrt{4}}{5-4} = \sqrt{5}-\sqrt{4}$

$\frac{1}{\sqrt{5}+\sqrt{6}}= \frac{1}{(\sqrt{6}+\sqrt{5})}\times \frac{(\sqrt{6}-\sqrt{5})}{(\sqrt{6}-\sqrt{5})}$

$= \frac{(\sqrt{6}-\sqrt{5})}{6-5}=\sqrt{6}-\sqrt{5}$

$\frac{1}{\sqrt{6}+\sqrt{7}}=\frac{1}{\sqrt{7}+\sqrt{6}} \times \frac{(\sqrt{7}-\sqrt{6})}{(\sqrt{7}-\sqrt{6})}=\frac{\sqrt{7}-\sqrt{6}}{7-6}$

$= \sqrt{7}-\sqrt{6}$

$\frac{1}{\sqrt{7}+\sqrt{8}}=\frac{1}{(\sqrt{8}+\sqrt{7})}\times \frac{(\sqrt{8}-\sqrt{7})}{(\sqrt{8}-\sqrt{7})} =\frac{(\sqrt{8}-\sqrt{7})}{8-7}$

$= \sqrt{8}-\sqrt{7}$

$\frac{1}{\sqrt{8}+\sqrt{9}}=\frac{1}{\sqrt{9}+\sqrt{8}}\times \frac{(\sqrt{9}-\sqrt{8})}{(\sqrt{9}-\sqrt{8})}=\frac{\sqrt{9}-\sqrt{8}}{9-8}$

$= \sqrt{9}-\sqrt{8}$

Now LHS = $\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}$

$=-\sqrt{4}+\sqrt{9}$

$=-2+3=1\: \: RHS$

Thus LHS =  RHS

## if  $x =3-2\sqrt{2}$  then find the value of  $x^{4}-\frac{1}{x^{4}}$

Given $x\: = 3-2\sqrt{2}$

$\frac{1}{x}= \frac{1}{(3-2\sqrt{2})}\times \frac{(3+ 2\sqrt{2})}{(3+2\sqrt{2})}\: \: (\: rationalizing\: )$

$\frac{1}{x}= \frac{(3+2\sqrt{2})}{9-8}= 3+2\sqrt{2}$

$\frac{1}{x^{2}}= (3+2\sqrt{2})^{2}=9+8+12\sqrt{2}=17+12\sqrt{2}$

$x^{2}= (3-2\sqrt{2})^{2}=9+8-12\sqrt{2}=17-12\sqrt{2}$

Now , $x^{4}-\frac{1}{x^{4}}=\left ( x^{2}-\frac{1}{x^{2}} \right )\left ( x^{2}+\frac{1}{x^{2}} \right )$

$=\left [ (17-12\sqrt{2})-(17+12\sqrt{2}) \right ] \left [ (17-12\sqrt{2}) +(17+12\sqrt{2})\right ]$

$= 17-12\sqrt{2}-17-12\sqrt{2}\left ( 17-12\sqrt{2}+17+12\sqrt{2} \right )$

$=(-24\sqrt{2})\times 34=-816\sqrt{2}$

## If     $x =\frac{\sqrt{5}+1}{\sqrt{5}-1}\: \: and\: \: y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$   than find the value of   $x^{2}+y^{2}$

$x = \frac{\sqrt{5}+1}{\sqrt{5}-1}$

$x^{2}= \left [ \frac{\sqrt{5}+1}{\sqrt{5}-1} \right ]^{2}= \frac{6+2\sqrt{5}}{6-2\sqrt{5}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$y^{2}=\left [ \frac{\sqrt{5}-1}{\sqrt{5}+1} \right ]^{2}=\frac{6-2\sqrt{5}}{6+2\sqrt{5}}=\frac{3-\sqrt{5}}{3+\sqrt{5}}$

$x^{2}+y^{2}=\frac{3+\sqrt{5}}{3-\sqrt{5}}+\frac{3-\sqrt{5}}{3+\sqrt{5}}$

$x^{2}+y^{2}=\frac{(3+\sqrt{5})^{2}}{(3-\sqrt{5})}+\frac{(3-\sqrt{5})^{2}}{(3+\sqrt{5})}$

$= \frac{9+5+6\sqrt{5}+9+5-6\sqrt{5}}{9-5}$

$= \frac{28}{4}$

$\therefore x^{2}+y^{2} =7$

## if      $a =\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$    and    $b =\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$    find the value  of    $\frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}$

$a^{2}+ab+b^{2} =(a+b)^{2}-ab$

$a^{2}-ab+b^{2} =(a-b)^{2}+ab$

$a = \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$= \frac{(\sqrt{5}+\sqrt{2})^{2}}{5-2}$

$= \frac{5+2+2\sqrt{10}}{3}= \frac{7+2\sqrt{10}}{3}$

$b = \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{7-2\sqrt{10}}{3}$

$(a+b)^{2}-ab = \left [ \frac{7+2\sqrt{10}}{3}+\frac{7-2\sqrt{10}}{3} \right ]^{2}$$-\left [ \frac{7+2\sqrt{10}}{3} \right ]\left [ \frac{7-2\sqrt{10}}{3} \right ]$

$= \left ( \frac{14}{3} \right )^{2}-\left ( \frac{49-40}{9} \right )$

$= \frac{196}{9}-\frac{9}{9}$

$= \frac{187}{9}$

$(a-b)^{2}-ab = \left ( \frac{7+2\sqrt{10}}{3}+\frac{7-2\sqrt{10}}{3} \right )^{2}$$+ \left ( \frac{7+2\sqrt{10}}{3} \right )\left ( \frac{7-2\sqrt{10}}{3} \right )$

$= \left ( \frac{4\sqrt{10}}{3} \right )^{2}+\frac{49-40}{9}$

$= \frac{160}{9}+\frac{9}{9} =\frac{169}{9}$

$\therefore \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}= \frac{(a+b)^{2}-ab}{(a-b)^{2}+ab}$

$= \frac{-187/9}{169/9}$

$=\frac{187}{169}$

## Find the value of    $\frac{3^{30}+3^{29}+3^{28}}{3^{31}+3^{30}-3^{29}}$

$\frac{3^{30}+3^{29}+3^{28}}{3^{31}+3^{30}-3^{29}} = \frac{3^{38}(3^{2}+3^{1}+1)}{3^{29}(3^{2}+3^{1}-1)}$

$= \frac{(9+3+1)}{3(9+3-1)}$

$= \frac{13}{3\times 11}=\frac{13}{33}$

## Find the value of      $\frac{4}{(216)^{\frac{-2}{3}}}+\frac{1}{(216)^{\frac{-3}{4}}}+\frac{2}{(343)^{\frac{-1}{5}}}$

$\frac{4}{(216)^{\frac{-2}{3}}}+\frac{1}{(216)^{\frac{-3}{4}}}+\frac{2}{(343)^{\frac{-1}{5}}}$      $= \frac{4}{(6^{3})^{\frac{-2}{3}}}+\frac{1}{(4^{4})\frac{-3}{4}}+\frac{2}{(3^{5})^{\frac{-1}{5}}}$

$=\frac{4}{6^{-2}}+\frac{1}{4^{-3}}+\frac{2}{3^{-1}}$

$=\frac{4}{36^{-1}}+\frac{1}{64^{-1}}+\frac{2}{3^{-1}}$

$= 4\times 36+1\times 64+2\times 3$

$= 214$

## Show that   $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}= 1$

$\frac{1}{1+x^{a-b}} +\frac{1}{1+x^{b-a}}$   $=\frac{x^{b}}{x^{b}+x^{a}} + \frac{x^{a}}{x^{a}+x^{b}}$

$=\frac{x^{b}+x^{a}}{x^{a}+x^{b}}$

$Alternative Method$

$\frac{1}{1+x^{a b}}+\frac{1}{1+x ^{ba}}= \frac{1}{1+\frac{x^{a}}{x^{b}}} +\frac{1}{1+\frac{x^{b}}{x^{a}}}$

$=\frac{1}{\frac{x^{b}+x^{a}}{x^{b}}}+\frac{1}{\frac{x^{a}+x^{b}}{x^{a}}}$

$=\frac{x^{b}}{x^{b}+x^{a}}$ $+\frac{x^{a}}{x^{a}+x^{b}}$

$= \frac{(x^{b}+x^{a})}{(x^{a}+x^{b})}$

## Prove that    $\left ( \frac{x^{a^{2}}}{x^{b^{2}}} \right )^{\frac{1}{a+b}}= \left ( \frac{x^{b^{}2}}{x ^{c^{}2}} \right )^{\frac{1}{b+c}}.\left ( \frac{x^{c^{}2}}{x^{a^{}2}} \right )^{\frac{1}{c+a}}=1$

$LHS = \left ( \frac{x^{a^{}2}}{x^{b^{}2}} \right ) ^{\frac{1}{a+b}}.\left ( \frac{x^{b^{}2}}{x^{c^{}2}} \right )^{\frac{1}{b+c}}.\left ( \frac{x^{c^{}2}}{x^{a^{}2}} \right )^{\frac{1}{c+a}}$

$=\left ( x^{a^{}2-b^{}2} \right )^{\frac{1}{a+b}}.\left ( x^{b^{}2-c^{}2} \right )^{\frac{1}{b+c}}.\left ( x^{c^{}2-a^{}2} \right )^{\frac{1}{c+a}}$

$= x^{\frac{a^{}2-b^{}2}{a+b}}.x^{\frac{b^{}2-c^{}2}{a+b}}.x^{\frac{c^{}2-a^{}2}{c+a}}$

$= x^{a-b}. x^{b-c}.x^{c-a}$

$=x^{0}$

$= 1 = RHS$

## Find the co-ordinate of the point which lies on y-axis at a distance of 4 units in negative direction of y-axis. (A) (-4,0)              (B) (4,0) (c) (0,-4)              (D) (0,4)

The point on y- axis  has x-axis has x- coordinate 0.

Since it lies at a distance of 4 units in the negative direction of y-axis.

$\therefore$The point  is (0, -4)

## If  $x=3-2\sqrt{2}$  find the value of   $\sqrt{x}+\frac{1}{\sqrt{x}}.$

$x =3-2\sqrt{2}\Rightarrow \frac{1}{x} = 3+2\sqrt{2}$

$\left ( \sqrt{x}+\frac{1}{\sqrt{x}} \right )^{2} = 8$

$\sqrt{x}+\frac{1}{\sqrt{x}} =\pm 2\sqrt{2}$

$Alternative \: \: Method$

$x= 3-\sqrt{2}$

$\frac{1}{x}=\frac{1}{(3-2\sqrt{2})}\times \frac{(3+2\sqrt{2})}{(3+2\sqrt{2})}$

$= \frac{\left ( 3+2\sqrt{2} \right )}{9-8}= 3+2\sqrt{2}$

$\left ( \sqrt{x}+ \frac{1}{\sqrt{x}} \right )^{2}= x+\frac{1}{x}+2\times \sqrt{x}\times \frac{1}{\sqrt{x}}$

$= 3-2\sqrt{2}+3+2\sqrt{2}+2$

$\left ( \sqrt{x}+\frac{1}{\sqrt{x}} \right )^{2} = 8$

$\left ( \sqrt{x}+\frac{1}{\sqrt{x}} \right ) = \pm 2\sqrt{2}$

## (i)Find the six rational  numbers bbetween 3 and 4  (ii) Which mathematical concept is used in this problem (iii) Which value is depticted in this question

(i) We known that between two rational numbers  x and y such that x< y there is a rational number $\frac{x+y}{2}$.

ie,       $3< \frac{7}{2}< 4$

Now a rational number between 3 and   $\frac{7}{2}< 4$   is :

$\frac{1}{2}\left ( 3+\frac{7}{2} \right )=\frac{1}{2}\times \left ( \frac{6+7}{2} \right )=\frac{13}{4}$

A rational nummber berween  $\frac{7}{2}\: \: and \: \: 4 \: \: is$

$\frac{1}{2}\left ( \frac{7}{2}+4 \right )=\frac{1}{2}\times \left ( \frac{7+8}{2} \right )=\frac{15}{4}$

$3< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< 4$

Further a rational number between 3 and  $\frac{13}{4}\: \: is :$

$\frac{1}{2}\left ( 3+\frac{13}{4} \right )=\frac{1}{2}\left ( \frac{12+13}{4} \right )=\frac{25}{8}$

A rational number between  $\frac{15}{4}\: \: and\: \: 4\: \: is :$

$\frac{1}{2}\left ( \frac{15}{4}+4 \right )=\frac{1}{2}\times \frac{15+16}{4}=\frac{31}{8}$

A rational number between  $\frac{31}{8}\: \: and\: \: 4\: \: is$

$\frac{1}{2}\left ( \frac{31}{8}+4 \right ) = \frac{1}{2}\times \left ( \frac{31+32}{8} \right )=\frac{63}{16}$

$\therefore 3< \frac{25}{8}< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< \frac{31}{8}< \frac{63}{16}< 4$

Hence , six rational numbers between 3 and 4 are

$\frac{25}{8}, \frac{13}{4},\frac{7}{2}, \frac{15}{4}, \frac{31}{8}\: \: and\: \: \frac{63}{16}$

(ii) Number syatem

(iii) Rationality is always welcomed

## Two classmates Salma and Anil simplified two different expressions during te revision hour and explained to each other their simplifications. Salma explains simplification of   $\frac{\sqrt{2}}{\sqrt{5}+\sqrt{3}}$    and Anil explains simplification of   $\sqrt{28}+\sqrt{98}+\sqrt{147}$.  Write the both simplification. What value does it depict ?

Justify      $\frac{\sqrt{2}}{\sqrt{5}+\sqrt{3}}= \frac{\sqrt{2}}{\sqrt{5}+\sqrt{3}}\times \frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})}$

$= \frac{\sqrt{10}-\sqrt{6}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}= \frac{\sqrt{10}-\sqrt{6}}{5-3}$

$=\frac{\sqrt{10}-\sqrt{6}}{2}$

Again    $\sqrt{28}+ \sqrt{98}+\sqrt{147}$

$=\sqrt{2\times 2\times 7}+\sqrt{2\times 7\times 7}+\sqrt{3\times 7\times 7}$

$= 2\sqrt{7}+7\sqrt{2}+7\sqrt{3}$

Co-operative learning among classmates with out any gender and religious bias.

## Write the expression which represents a polynomial

$\sqrt{3}\: \: x^{2}-x-1$

Not defined

## Name the degree of the polynomial  -3x + 2.

Linear polynomial

## Write an exampleof a constant polynomial.

Constant polynomial is 7.

## What is  $x + \frac{1}{x}\: \: ?$

Not a polynomial

Binomial

## What is the degree of polynomial $\sqrt{3}\: \: ?$

Degree of a polynomial  $\sqrt{3}\: \: is\: \: 0\: \:$

## What is the degree of the polynomial  $(x^{3}+5) (4-x^{5})$  ?

Degree of  $x^{3}+5 = 3$

Degree of  $4-x^{5}=5$

Degree of   $(x^{3}-5)(4-x^{5})=3+5=8.$

## What is the zero of the zero polynomial ?

Every real number is a zero of the zero polynomial

## Write the number of zeroes in a cubic polynomial

No of the zeroes of cubic polynomial  = 3

## Find   $p(0)\: \: if \: \: p (y)= y^{2}-y+1.$

$p(y)=y^{2}-y+1$

$p(0)=0^{2}-0+1 =1$

## Write any polynomial in one variable

$\sqrt{2}\: \: x^{2}+3x \: \: or\: \: \sqrt{3}\: \: y^{2}+3y$

## In the expression   $x^{2}+\frac{\pi }{2}\: x-7$   what is the co-efficient of x ?

Co -efficient of x in expression   $x^{2}+\frac{\pi }{2}\: x-7\: \: is\: \: \frac{\pi }{2}$

## If -4 is a zero of the polynomial   $p(x)= x ^{2}+11x+k$   then calculate the value of k.

Given , $p(x )=x^{2}+11x+k$

Since -4 is a zero of polynomial

$p(-4)=0$

$(-4)^{2}+11\times (-4)+k =0$

$16-44+k =0$

$k = 28.$

## Write the zeroes of the polynomial  $p (x)= x(x-2)(x-3).$

For zeroes , put  $p (x )=0$

$x(x-2) (x-3 )=0$

$x =0, 2,3$

## if   $p(x)= x^{2}-3x+2$   then what is the value of   $p(0)+p(2)\: \: ?$

Putting $x =0$

$p (0)= 0-3\times 0+2=2$

$x = 2$

$p(2)=2^{2}-3\times 2+2$

$=4-6+2=0$

Thus   $p(0)+p(2)= 2+0=2$

## Clasify the following as linear , quadratic and cubic polynomials  (a)  $x^{2}+x$       (b)   $x -x^{3}$  (c) 1+x            (d)   $7x^{3}$

linear polynomial   $\rightarrow 1+x,$   degree 1

Quadratic Polynomial   $\rightarrow x^{2}+x$    degree 2

Cubic Polynomial   $\rightarrow x - x^{3}, 7x^{3}$   degree 3

## Find the value of polynomial     $p(x)= x^{3}-3x^{2}-2x+6 \: \: at\: \: \: x = \sqrt{12}$

$p(x)= x^{3}}-3x^{2}-2x +6$

$p(\sqrt{2})= (\sqrt{2})^{3}-3(\sqrt{2})^{2}-2 (\sqrt{2})+6$

$= 2\sqrt{2}-6-2\sqrt{2}+6$

$= 0$

## if y = 2  and  y = 0  are the zeroes of the polynomial  $f(y )=2y^{3}-5y^{2}+ay+b$  find the value of a and b

$f(y)=2y^{3}-5y^{2}+ay+b$

$f(2)=2(2)^{3}-5(2)^{2}+a(2)+b =0$

$16-20+2a+b =0$

$2a +b = 4$ .......................(1)

$f (0)= b=0$

$From\: \: (i)\: \: 2a +0=4$

$a=2$

$a=2, b= 0$

## If  $f(x)= 3x+5,$    evaluate   $f(7)-f(5)$

$f(x)=3x+5$

$f(7)=3\times 7+5=26$

$f(5)=3\times 5+5=20$

$f(7)-f(5)=26-20=6.$

## Find the value of the polynomial   $x^{2}-3x+6 \: \: at$ (1) $x=\sqrt{2}$                (2)   $x=3$

$p(x)=x^{2}-3x+6$

(1) When  $x =\sqrt{2}$

then   $p(\sqrt{2})=(\sqrt{2})^{2}-3\times \sqrt{2}+6$

$=2-3\sqrt{2}+6$

$=8-3\sqrt{2}$

(2) When $x=3$

then   $p(3)=3^{2}-3\times 3+6$

$=9-9+6$

$=6$

## if   $f(x)=x^{4}-4x^{3}+3x^{2}-2x+1$   then find whether   $f(0)\times f(-1)=f(2)$

$f(x)\: \:= x^{4}-4x^{3}+3x^{2}-2x+1$

$f(0)=1$

$f(-1)= (-1)^{4}-4(-1)^{3}+3(-1)^{2}-2(-1)+1$

$=1+4+3+2+1=11$

$f(2)=(2)^{4}-4(2)^{3}+3(2)^{2}-2(2)+1$

$=16-32+12-4+1$

$=29-36=-7$

$\therefore f (0)\times f(-1)=11\neq f(2)$

## If $f(x) = x^{3}-3x^{2}+3x-4\: \: find \: \: f (2)+f(-2)+f(0).$

$f(x)=x^{3}-3x^{2}+3x-4$

$f(2)=(2)^{3}-3(2)^{2}+3(2)-4$

$=8-12+6-4$

$f(2)=-2$

$f(-2)=(-2)^{3}-3(-2)^{2}+3(-2)-4$

$=-8-12-6-4$

$f(-2)=-30$

$f(0)=-4$

$\therefore f(2)+f(-2)+f(0)=-2-30-4=-36$

## If   $f(x)=5x^{2}-4x+5$   find  $f(1)+f(-1)+f(0).$

$f(x)= 5x^{2}-4x+5$

$f(1)=5-4+5$

$=6$

$f(-1)=5(-1)^{2}-4(-1)+5$

$=5+4+5$

$=14$

$f(0)=5$

$\therefore f(1)+f(-1)+f(0)=6+14+5=25$

## $if\: \: \: f(x)=x^{2}-5x+7,\: \: \: evaluate\: \: \: f(2)-f(-1)+f\left ( \frac{1}{3} \right )$

$f(x)= x^{2}-5x+7$

Then  $f (2)=2^{2}-5\times 2+7=1$

$f (-1)=(-1)^{2}-5(-1)+7=13$

$f \left ( \frac{1}{3} \right )=\left ( \frac{1}{3} \right )^{2}-5\left ( \frac{1}{3} \right )+7=\frac{49}{9}$

$f(2)-f(-1)+f \left ( \frac{1}{3} \right )=1-13+\frac{49}{9}=\frac{-59}{9}$

## If  $p(x)=x^{3}+3x^{2}-2x+4$  then find the value of  $p(2)+p(-2)-p(0).$

$p(x)=x^{3}+3x^{2}-2x+4$

$p(2)=(2)^{3}+3(2)^{2}-2(2)+4=8+12-4+4 =20$

$p(-2)=(-2)^{3}+3(-2)^{2}-2(-2)+4$

$=-8+12+4+4=12$

$p(0)=4$

$p(2)+p(-2)-p(0)=20+12-4=28.$

## On dividing   $5y^{3}-2y^{2}-7y +1$   by y what remainder do we get ?

Let , $p (y) = 5y^{3}-2y^{2} -7y +1$

then remainder = $p(0)= 0-0-0+1=1$

## If   $x^{11}+101$   is divided by x+1 then what remainder do we get ?

let, $f(x)= x^{11}+101$

Put  $x+1= 0$

$\Rightarrow x = -1$

Then remainder is :

$f(-1)=(-1)^{11}+101$

$= -1+101=100.$

## Find the remainder when   $x^{3}+x^{2}+x+1$  is divided by   $x-\frac{1}{2}$   using remainder theorem .

$p(x) = x^{3}+x^{2}+x+1$

$x-\frac{1}{2}=0$

$\Rightarrow x = \frac{1}{2} \: \: in\: \: p (x)$

$Remainder = p \left ( \frac{1}{2} \right )$

$=\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )+1$

$=\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1$

$=\frac{1+2+4+8}{8}$

$=\frac{15}{8}.$

## Find the remainder when  $x^{3}+6x-ax^{2}-a$   is divided by x - a

Here ,  $p(x)=x^{3}-ax^{2}+6x-a$

and the zero of x - a is a.

so,   $p(a)= a^{3}-a.a^{2}+6a-a =5a$

So, by the remainder theorem 5a is the remainder when  $x^{3}-ax^{2}+6x-a$   is divided by x-a

## Find the remainder when the polynomial  $f(x) =4x^{3}-12x^{2}+14x-3$   is divided by  $(2x-1).$

2x -1 = 0

$x= \frac{1}{2}$

$By\: \: remainder\: \: theorem\: ,\:$

$if\: \: f(x)\: \: \: \: is\: \:\: divided\: \:\: \: by\: \: 2x -1\: \: the\: \: remainder\: \: is\: \: f\left( \frac{1}{2} \right )$

$\therefore f\left ( \frac{1}{2} \right )= 4\left ( \frac{1}{2} \right )^{3}-12\left ( \frac{1}{2} \right )^{2}+14\left ( \frac{1}{2} \right )-3$

$= 4\times \frac{1}{8}-12\times \frac{1}{4}+14\times \frac{1}{2}-3$

$=\frac{1}{2}-3+7-3$

$=\frac{1}{2}+1$

$=\frac{3}{2}$

Hence required remainder is   $\frac{3}{2}.$

## By actual division, find the quotient and remainder when  $p(x)= 2x^{3}+3x^{2}-9x+4$  is divided by 2x-1.

$p(x)= 2x^{3}+3x^{2}-9x+4$

$Quotient\: \: = \: \: x^{2}+2x-\frac{7}{2}$

$Remainder\: \: =\: \: \frac{1}{2}$

## Using remainder theorem, factorize  $6x^{3}-25x^{2}+32x-12.$

Factors of   $12=(\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,)$

$p(x)=6x^{3}-25x^{2}+32x-12$

$p(2)= 6(2)^{3}-25(2)^{2}+32\times 2-12$

$= 48-100+64-12$

$=112-112=0$

$\therefore x = 2$  is a zero of  $p(x )\: \: or\: \: (x-2)$  is a factor  $p(x)$

$6x^{3}-25x^{2}+32x-12$

$= 6x^{2}(x-2)-13x(x-2)+6x(x-2)$

$= (x-2)(6x^{2}-13+6)$

$=(x-2)(6x^{2}-9x-4x+6)$

$=(x-2)\left [ 3x(2x-3) -2(2x-3)\right ]$

$=(x-2)(2x-3)(3x-2).$

## Polynomial  $3x^{3}-5x^{2}+kx-2\: \: and\: \: -x^{3}-x^{2}+7x+k$  leave the same remainder when divided by $x+2$  find the value at k.

Let ,  $p(x)=3x^{3}-5x^{2}+kx-2$  and

$q(x)= -x^{3}-x^{2}+7x+k$

Put   $x+2 = 0$

or   $x= -2\: in \: \: p(x) \: \: and\: \: q(x)$

$p(-2)=3(-2)^{3}-5(-2)^{2}+k(-2)-2$

$=-24-20-2k-2$

$=-2k-46$

$q(-2)=-(-2)^{3}-(-2)^{2}+7(-2)+k$

$=8-4-14+k$

$=-10+k$

$\therefore p (x)$    and    $q (x)$   leave the same remainder when divided by x+2.

$\Rightarrow -2k-46=k-10$

$\Rightarrow -3k = 36$

$\therefore k= -12$

## The polynomial  $ax^{3}+3x^{2}-13$  and  $2x^{3}-5x+a$  leave the same remainder in each case when divided by (x-2). Find the value of a

Let,      $p(x)= ax^{3}+3x^{2}-13$

$g(x)= 2x^{3}-5x+a$

$R_{1} = p(2)$

$R_{2}= g(2)$

$p(2 )= a(2)^{3}+3(2)^{2}-13$

$R_{1}= 8a+12-13$

$= 8a-1$

$R_{2}= 2(2)^{3}-5(2)+a$

$=16-10+a$

$\Rightarrow R_{2}=6+a$

$\therefore R_{1}=R_{2}$

$8a-1=6+a$

$7a=7$

$\therefore a =1$

## Find the value of 'a' if remainder is same when polynomial  $p(x)=x^{3}+8x^{2}+17x+ax$  is divided by  $(x+2)$  and  $(x+1)$

$p(x)=x^{3}+8x^{2}+17x+ax$

$p(x)$ leave the same remainder when divided by  $(x+2)$  and  $(x+1)$.

$p(-2)= (-2)^{3}+8(-2)^{2}+17(-2)+a(-2)$

$=-8+32-34-2a$

$=-10-2a$

$p(-1)= (-1)^{3}+8(-1)^{2}+17(-1)+a (-1)$

$=-1+8-17-a$

$=-10-a$

$\because$  Remainders are equal

so, $-10-2a = -10-a$

$\Rightarrow - a =0$

$\therefore a = 0$

## If the polynomial   $ax^{3} +4x^{2}+3x-4$    and   $x^{3}-4x+a$   leave the same remainder when divided by (x - 3)  Find the value of a.

Let ,      $p(x)= ax^{3}+4x^{2}+3x-4$

and       $q(x)= x^{3}-4x+a$

Put     $x-3$ = 0  or  $x=3$  in $p(x )$ and $q(x )$

$p(3)= a(3)^{3}+4(3)^{2}+3\times 3-4$

$= 27a +36+9-4$

$=27a+41$

$q(3)= (3)^{3}-4(3)+a$

$=27-12+a$

$= 15+a$

According to the  question

$p(3)=q(3)$

$27a+41= 15+a$

$\Rightarrow 27a-a = 15-41$

$\Rightarrow 26a= -26$

$\Rightarrow a=-1$.

## What must be substracted from  $x^{4}+1$  so that   $x^{4}+1$  is exactly divisible by (x-1). write the resultant polynomial which is exactly divisible by (x-1).

$p(x) =x^{4}+1$

Put, $x-1 = 0$  or  x = 1 in $p(x)$

$p(1)= (1)^{4}+1$

$= 1+1 =2$

Hence , 2 must be substracted from $x^{4}+1$ so that it is exactly divisible b (x-1).

Resultant polynomial to be divisible by (x-1)

$= x^{4}+1-2$

$=x^{4}-1$.

## Find the value of p if the polynomial   $p(x)= x^{4}-2x^{3}+3x^{2}-px+3p-7$   when divided by   $(x+1)$  leaves  the remainder 19. Also find the remainder when p (x) is divided by x+2.

$p(x)=x^{4}-2x^{3}+3x^{2}-px +3p-7$

$Put, \: x+1 = 0 \: \: or\: \: x = -1 \: \: in \: \: p(x), \: we \: \: get$

$p(-1)= (-1)^{4}-2(-1)^{3}+3(-1)^{2}$-$-p(-1)+3p-7=19$

$\Rightarrow 1+2+3+p+3p-7 =19$

$\Rightarrow 4 p -1 = 19$

$\Rightarrow 4p=20$

$\therefore$ $\Rightarrow p=5$

$The\: \: polynomial\: \: p(x)=x^{4}-2x^{3}+3x^{2}-5x+15-7$

$= x^{4}-2x^{3}+3x^{2}-5x+8$

$Put\: \: x+2 =0\: \: or\: \: x=-2\: \: in\: \: p(x)$

$p(-2)=(-2)^{4}-2(-2)^{3}+3(-2)^{2}-5(-2)+8$

$=16+16+12+10+8$

$=62.$

## The  polynomials  $\: \: ax^{3}-3x^{2}+4 \: \: and\: \: 2x^{3}-5x+a \: \:$   when  divided  by (X - 2) leave  the remainders  p  and  q  respectively  . $If\: \: p-2q =4\: \:$  find the  value  of  a.

Let , $f (x)=ax^{3}-3x^{2}+4$

$g(x)=2x^{3}-5x+a$

When divided by  $(x-2)$

$f(2)=p$    and    $g(2)=q$

$f(2)= a\times 2^{3}-3\times 2^{2}+4$

$\Rightarrow p = 8a-12+4$

$\Rightarrow p = 8a-8$

$g (2)=2\times 2^{3}-5\times 2+a$

$\Rightarrow q =16-10+a$

$\Rightarrow q =6+a$

$p-2q = 4g$

$\Rightarrow 8a-8-12-2a =4$

$6a-20=4$

$a=4$

## If   $f(x)=x^{4}-2x^{3}+3x^{2}-ax+b$   is a polynomial such that when it is divided by $x-1$ and $x+1$ the remaniders are 5 and 19 respectively . Determine the remainders when  $f(x)$  is divided by $x-2.$

$f(x)=x^{4}-2x^{3}+3x^{2}-ax+b$

put x-1    or   x=1  in  $f(x)$   we get.

$\Rightarrow f(1)=(1)^{4}-2(1)^{3}+3(1)^{2}-a\times 1+b$

$\Rightarrow 5=1-2+3-a+b$

$\Rightarrow 5=2-a+b$

$\Rightarrow a-b =2-5$

$\Rightarrow a-b = -3$  ..........................(1)

Again put x+1  = 0    or   x = -1 in  f(x)  we get.

$f(-1)= (-1)^{4}-2(-1)^{3}+3(-1)^{2}-a(-1)+b$

$\Rightarrow 19 = 1+2+3+a+b$

$\Rightarrow 19 -6= a+b$

$\Rightarrow a+b = 13$  ...........................(2)

$2a = 10$

$a=5$

By equation (2)

$5+b=13$

$\therefore b =8$

$\therefore f(x)= x^{4}-2x^{3}+3x^{2}-5x+8$

Again Put, $x-2 = 0$  or  $x=2$  in  $f(x)$

$f(2)=(2)^{4}-2(2)^{3}+3(2)^{2}-5\times 2+8$

$=16-16+12-10+8$

$=20-10=10$

## The polynomials  $bx^{3}+3x^{2}-3\: \: and\: \: 2x^{3}-5x+b$  when divided by $x-4$ leave the remainders $R_{1}$ and $R_{2}$ respectively .  Find the value of b if   $2R_{1}-R_{2} =0$

Let  $p(x) = bx^{3}+3x^{2}-3$

Put  $x-4 = 0$  or  $x=4$  in $p(x)$

We get  $p(4)= b (4)^{3}+3(4)^{2}-3$

$=R_{1}$

$64 b +48-3 = R_{1}$

$64 b +45 = R_{1}$

Let, $q (x) = 2x^{3}-5x+b$

Again  put x - 4 = 0 or x = 4 in $q (x)$

$q(4)= 2(4)^{3}-5(4)+b=R_{2}$

$=128-20+b = R_{2}$

$108+b = R_{2}$

Given that $2R_{1}-R_{2}= 0$

$\Rightarrow 2(64b +45)-(108+b) =0$

$\Rightarrow 128b +90-108-b =0$

$\Rightarrow 127b-18=0$

$b = \frac{18}{127}$

## Factorize :$x^{2}3x$

$x^{2}-3x = x(x-3)$

## Factorize $12a^{2}b -6ab^{2}.$

$12a^{2}b-6ab^{2}(2a-b)$

## If f$f(x)$ be a polynomial such that $f \left (- \frac{1}{3} \right )=0$ then calculate one factor of $f(x)$

Since, $f\left ( -\frac{1}{3} \right )= 0$

$\therefore - \frac{1}{3}$ is a polynomial $f(x)$

So, $x+\frac{1}{3}$ or $3x+1$ is a factor of $f(x)$.

## Find the value of m, if x+4 is a factor of the polynomial $x^{2}+3x+m.$

$x+4$ is a factor $x^{2}+3x+m$$=p(x)$

$\Rightarrow p(-4)=0$

$\Rightarrow 16-12+m=0$

$\Rightarrow m= -4$

## Find the value of k if x-2 is a factor of $p(x)=2x^{2}+3x-k$

x-2 is a factor of p(x) then p(2)=0

$p(2)=2\times (2)^{2}+3\times 2-k =0$

$8+6-k =0$

$k=14.$

## Find the value of k if 2x-1 is a factor of the polynomial $6x^{2}+kx-2$.

2x-1 is a  factor of $p(x)= 6x^{2}+kx-2$

$p \left ( \frac{1}{2} \right )=0$

$6.\frac{1}{4}+k.\frac{1}{2}-2 =0$

$k=1$

## Write the factors of $a^{7}+ab^{6}.$

Since , $a^{7}+ab^{6}= a(a^{6}+b^{6})$

$\therefore$ Factors are a and  $(a^{6}+b^{6})$

## Factorize $20x^{2}-9x+1$

$20x^{2}-9x+1=20x^{2}-5x-4x+1$

$=5x(4x-1)-1(4x-1)$

$= (4x-1)(5x-1)$

## Factorize : $6-x+x^{2}$

$6x-x-x^{2}=6-3x+2x-x^{2}$

$=3(2-x)+x(2-x)$

$=(2-x)(3+x)$

## Factorize$8y^{3}-125x^{3}.$

$8y^{3}-125x^{3}= (2y)^{3}-(5x)^{3}$

$=(2y-5x)(4y^{2}+10xy+25x^{2})$

## write the factors of $a^{3}-1$

$a^{3}-1=(a-1)(a^{2}+1+a\times 1)$

Then factors of $(a^{3}-1)$ are$(a-1)$ and $(a^{2}+1+a)$

## Write the factors of polynomial $4x^{2}+y^{2}+4xy+8x+4y+4$.

$4x^{2}+y^{2}+4xy+8x+4y+4$

$=(2x)^{2}+(y)^{2}+(2)^{2}+2\times 2x+y+2\times 2x\times 2+2\times y\times 2$

$=(2x+y+2)^{2}.$

$Factots is 2x+y+2$

## Write the factors of the polynomial : $x^{2}+5\sqrt{2x}+12.$

$x^{2}+5\sqrt{2x}+12= x^{2}+3\sqrt{2x}+2\sqrt{2x}+12$

$=x x(x+3\sqrt{2})+2\sqrt{2}(x+3\sqrt{2})$

$= (x+3\sqrt{2})(x+2\sqrt{2})$

$Factors are x+3 and x+2\sqrt{2}.$

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## Write one factor of $(x+1)^{3}-(x+1)$

$(x+1)^{3}-(x+1)= (x+1)\left [ (x+1)^{2}-1 \right ]$

One Factor $(x+1)^{3}-(x+1)$ is $(x+1).$

## Find the value of k so that polynomial $x^{3}+3x^{2}-kx-3$ has one factor as x+3.

$f(x)=x^{3}+3x^{2}-kx-3$

$(x+3)$ is a factor of $f(x)= x^{3}+3x^{2}-kx-3$

$\Rightarrow f(-3)=0$

$\Rightarrow (-3)^{3}+3(-3)^{2}-k(-3)-3=0$

$\Rightarrow -27+27+3k-3=0$

$\Rightarrow 3k-3=0$

$\Rightarrow k= 1$

## Find the value of k if  $x-2$ is a factor of $p(x)=x^{2}+kx+2k$

$(x-2)$ is  a factor of $f(x)= x^{2}+kx+2k$

$f(2) = 0$

$\Rightarrow (2)^{2}+k(2)+2k =0$

$\Rightarrow 4+2k+2=0$

$4+4k =0$

$\Rightarrow k =-1$

## Find the value of 'a' for which (x-1) is factor of the polynomial $a^{2}x^{3}-4ax+4a-1.$

(x-1) is a factor of $f(x) = a^{2}x^{3}-4ax+4a-1$

$f(1)=0$

$\Rightarrow a^{2}(1)^{3}-4a(1)+4a-1 = 0$

$\Rightarrow a^{2}-4a+4a-1=0$

$\Rightarrow a^{2}-1=0$

$\Rightarrow a=\pm 1$

## For what value of k (x+1) is a factor of $p(x)= kx^{2}-x-4$

Given , $p(x)= kx^{2}-x-4$

(x+1) is a factor of p(x) then

$p(-1)=0$.

$\therefore k (-1)^{2}-(-1)-4=0$

$\Rightarrow k+1-4=0$

$\therefore k=3$

## Show that (x-1) is a factor of the polynomial $f(x)= 2x^{3}-3x^{2}+7x-6$

$f(x)= 2x^{3}-3x^{2}+7x-6$

$x-1= 0$ or $x = 1$  in  $f(x)$

Thus,  $f(1)= 2\times 1^{3}-3\times 1^{2}+7\times 1-6$

$= 2-3+7-6=0$

Hence  (x-1)  is a factor of f(x)

## For what value of k is the polynomial $p(x)=2x^{3}$$-kx^{2}+3x+10$ exactly divisible by (x+2)

(x+2) is a factor of $p(x)=2x^{3}$$-kx^{2}+3x+10$

$\Rightarrow p(-2)=0$

$\Rightarrow 2(-2)^{3}-k(-2)^{2}+3(-2)+10=0$

$\Rightarrow -16-4k-6+10=0$

$k =-3$

## Find the value of 'k' if (x-1) is a factor of $p(x)=2x^{2}+kx+\sqrt{2}$

$p(x)=2x^{2}+kx+\sqrt{2}$

$(x-1)$is a factor of $p(x)$ then $p(1)=0$

$\therefore 2(1)^{2}+k(1)+\sqrt{2}=0$

$\Rightarrow 2+k+\sqrt{2}=0$

$\therefore k = -2 - \sqrt{2}$

## Factorize : $x^{3}-3x^{2}-9x-5.$

Let, $p(x)=x^{3}-3x^{2}-9x-5$

$p(1)=-1-3+9-5=0$

$Since (x+1)$ is a factor of $x^{3}-3x^{2}-9x-5$

$\therefore (x^{3}-3x^{2}-9x-5)=(x+1)(x^{2}-4x-5)$

$x^{2}-4x-5=x^{2}-5x+x-5$

$=x(x-5)+1(x+5)$

$=(x+1)(x-5)$

Fatots are : $(x+1)(x-5)$

## If x-a is the factor of $3x^{2}-mx-nx$ then prove that $a=\frac{m+n}{3}$

$p(x)= 3x^{2}-mx-nx$

If (x-a) is a factor of p(x) then

$p(a)=0$

$\Rightarrow 3(a)^{2}-m\times a-n\times a=0$

$\Rightarrow a\left [ 3a-m-n \right ]=0, a\neq 0$

$\Rightarrow 3a-m-n =0$

$\therefore a = \frac{m+n}{3}.$

## Find the value of a for which (x-a) is   a factor of the polynomial $x^{6}-ax^{5}+x^{4}-ax^{3}+3x-a+2$

Let $p(x)=x^{6}-ax^{5}+x^{4}-ax^{3}+3x-a+2$

(x-a) is a  factor of the polynomial p(x) then p (a)=0

$\Rightarrow a^{6}-a\times a^{5}+a^{4}-a\times a^{3}+3\times a-a+2$ = 0

$\Rightarrow a^{6}-a^{6}+a^{4}-a^{4}+3a-a+2 =0$

$\Rightarrow 2a = -2\Rightarrow a =-1.$

## If (x-2) and $\left ( x-\frac{1}{2} \right )$ are factors of $p(x)^{2}+5x+r$ then show that p=r .

Let , $f (x)=px^{2}+5x+r$

$f (2)=0 and f\left ( \frac{1}{2} \right )=0$

$f(2)= 0\Rightarrow 4 p +10+r=0$

$4p+r = -10$

$f\left ( \frac{1}{2} \right )=0\Rightarrow \frac{p}{4}+\frac{5}{2}+r =0$

$\Rightarrow p+10+4r =0$

$\Rightarrow p+4r=-10$

(1)=(2) as both are equal to -10

$\therefore p+4r = 4p +r$

$\Rightarrow 4r-r=4p-p$

$\Rightarrow 3r = 3p$

$\Rightarrow r=p$

## Using factor theorem show that (m-n), (n-p) and (p-m) are factors of $m(n^{2}-p^{2})+n(p^{2}-m^{2})+p(m^{2}-n^{2})$

Let $p= m(n^{2}-p^{2})+n(p^{2}-m^{2})+p(m^{2}-n^{2})$

$p(m=n)=n(n^{2}-p^{2})+n(p^{2}-n^{2})+p(n^{2}-n^{2})$

$= n (n^{2}-p^{2})-n(n^{2}-p^{2})+0$

=0

$m-n is a factor of p$

Similarly $p(n=p)=0$ &$p(p=m)=0$

n-p is a factor of  p and p -m is a factor of p.

## Factroize : $9x^{3}-3x^{2}-5x-1.$

$p(x)= 9x^{3}-3x^{2}-5x-1.$

Factor of $1=\pm 1$

$p(1)=9-3-5-1=0$

$\Rightarrow x-1$ is a factor of $p(x)$

$\therefore 9x^{3}-3x^{2}-5x-1=9x^{2}(x-1)+6x(x-1)+1(x-1)$

$=(x-1)(9x^{2}+6x+1)$

$=(x-1)(3x+1)^{2}$

$\therefore p(x)=(x-1)(3x+1)(3x+1)$

Factors of   5

## Factorize : $x^{3}-2x^{2}-5x-6$

factor of  6 =  $6=(\pm 1,\pm 2,\pm 3,\pm 6)$

$p(x)= x^{3}+2x^{2}-5x-6$

$p(-1)= (-1)^{3}+2(-1)^{2}-5(-1)-6$

$=-1+2+5-6$

$=7-7=0$

$\because x = -1$ is zero of p (x) of (x+1) is a factor of p (x)

$\therefore x^{3}+2x^{2}-5x-6$

$=x^{2}(x+1)+x(x+1)-6(x+1)$

$= (x+1)[x^{2}+x-6]$

$= (x+1)[x(x+3)-2(x+3)]$

$= (x+1)(x+3)(x-2)$

## factorize $x^{3}+13x^{2}+32x+20$

(x+2)  is a factor of $x^{3}+13x^{2}+32x +20$

$\because [p(-2)=0]$

$x^{3}+13x^{2}+32x+20 = (x+2)(x^{2}+11x+10)$

$\rightarrow x^{2}+ 11x+10= x^{2}+10x+x+10$

$= x (x+10)+1(x+10)$

$= (x+1) (x+10)$

Factors are: (x+2)   (x+1)  (x+10)

Alternative method

Factors  of  20 =$(\pm 1,\pm 2,\pm 3,\pm 5,\pm 10,\pm 20)$

$p(x)=x^{3}+13x^{2}+32x+20$

$p(-1)= (-1)^{3}+13(-1)^{2}+32(-1)+20$

$=-1+13+32+20$

$33-33=0$

$\therefore x = -1$ is a zero of p(x), and (x+1) is a factor of p(x)

Then

$x^{3}+13x^{2}+32x+20$

$= x^{2}(x+1)+12x(x+1)+20 (x+1)$

$= (x+1)(x^{2}+12x+20)$

$= (x+1) [x (x+10)+2 (x+10)]$

$= (x+1)(x+2) (x+10)$

## Verify if 1 and  - 3  are zeroes of the polynomial $3x^{3}-5x^{2}-11x+3.$  if yes, factorize the polynomial.

$p (x)= 3x^{3}+5x^{2}-11x +3$

$p (1)= 3+5-11+3 =0$

$\therefore 1$ is a zero of p(x)

$p(-3)= -81+45+33+3 = 0$

-3 is a zero of p(x)

$(x-1) (x+3) = x^{2}+2x -3$ is a factor of p(x)

$\frac{p(x)}{x^{2}+2x-3} = 3x-1$, when  we divide physically

Hence, $p (x) = (x-1) (x+3) ( 3x-1)$

## Verify that (x-1), (x-2) and (2x+1) are the factors of the polynomial $2x^{3}-5x^{2}-x+2$

Let,  $p (x) = 2x^{3}-5x^{2}+x+2$

Put x = 1 in p (x)

$p (1) = 2(1)^{3}-5(1)^{2}+1+2$

$=2-5+1+2=0$

Hence, (x-1) is a factor of p (x).

Put x =  2 in p(x)

$p(2)=2(2)^{3}-5(2)^{2}+2+2 =16-20+2+2 =0$

hence (x-2) is a factor of p(x)

Put, $x = -\frac{1}{2}$ in p (x)

$p \left ( -\frac{1}{2} \right )=2$

## Verify that (x-1), (x-2) and (2x+1) arec the factors of the polynomial $2x^{3}-5x^{2}+x+2$

Let ,  $p (x)= 2x^{3}-5x^{2}+x+2$

Put x = 1 in p (x)

$p (1)= 2 (1)^{3}-5(1)^{2}+1+2$

$=2-5+1+2 =0$

Hence (x-1) is a factor of p (x)

Put x = 2

$p(2)= 2(2)^{3}-5(2)^{2}+2+2$

$=16-20+2+2=0$

Hence (x-2) is a  factor  of p(x)

Put $x = -\frac{1}{2}$ in p (x)

$x = -\left ( \frac{1}{2} \right ) = 2\times \left ( -\frac{1}{2} \right )^{3}-5 \left ( -\frac{1}{2} \right )^{2}+\left ( -\frac{1}{2} \right )+2$

$2\times \left ( -\frac{1}{8}\right )-5\times \frac{1}{4}-\frac{1}{2}+\frac{2}{1}$

$=\frac{-1-5-2+8}{4} = \frac{0}{4} = 0$

Hence $\left ( x+\frac{1}{2} \right )$ or  (2x+1) is a factor of p (x)

## Using factor theorem, find the value of a if $2x^{4}-ax^{3}+4x^{2}-x+2$ is divisible be 2x+1.

$[p (x) = 2x^{4}-ax^{3}+4x^{2}-x+2]$

If  (2x+1)is a factor of p(x) then 2x+1= 0,  $[x = \frac{-1}{2}]$

is a zero  of the polynomial   p(x)

so,  $[p \left ( -\frac{1}{2} \right ) = 0]$

$p\left ( -\frac{1}{2} \right )=2\times \left ( \frac{-1}{2} \right )^{4}-a \left ( \frac{-1}{a} \right )^{3} +4\left ( \frac{-1}{4} \right )^{3}-\left ( \frac{-1}{2} \right )+2$

$=0$

$\Rightarrow 2\times \frac{1}{16}-a\left ( \frac{-1}{8} \right )+4\left ( \frac{1}{4} \right )-\left ( \frac{-1}{2} \right )+2=0$

$\Rightarrow \frac{1}{8}+\frac{a}{8} +1+\frac{1}{2}+2=0$

$\Rightarrow \frac{29}{8}+\frac{a}{8} = 0$

$\therefore a = -29$

## Find the value  of p for which the polynomial$x^{3}+4x^{2}-px+8$exactly divisible by  x-2 Hence factorize the polynomial.

$p (x) = x^{3}+4x^{2}-px+8$

p (x) is exactly divisible be x- 2

$\therefore p (2) = 0$

$\rightarrow p (2)=0$

$\rightarrow p (2)=(2)^{3}+4(2)^{2}-p (2)+8 = 0$

$\rightarrow 8+16-2p+8 = 0$

$\rightarrow 32-2p =0$

$\therefore p = 16$

$\therefore p(x) = x^{3}+4x^{2}-16x+8$

Then,

$x^{3}+4x^{2}-16x+8 = x^{2}(x-2)+6x(x-2)-4(x-2)$

$=(x-2)(x^{2}+6x-4)$

## With out actual division show that $f (x)=2x^{4}-6x^{3}+3x^{2}+3x-2$ is exactly divisible by $x^{2}-3x+2.$

Let, $g(x) = x^{2}-3x+2$

$=x^{2}-2x-x+2$

$=x (x-2) - 1 (x-2)$

$= (x-2) (x-1)$

Zero of x -2 is 2 as  x-2=0  $\rightarrow$ x = 2

Zero of x -1 is 1 as  x-1=0    $\rightarrow$x = 1

Now,

$f (x)=2x^{4}-6x^{3}+3x^{2}+3x-2$

$f (2)=2(2)^{4}-6(2)^{3}+3(2)^{2}+3(2)-2$

$=32-48+12+6-2 =0$

$f (1)=2(1)^{4}-6(1)^{3}+3(1)^{2}+3(1)-2$

$=2-6+3+3-2 =0$

$\rightarrow (x-1)$ and (x-2) are the factors of f (x)

$\therefore f (x)$is exactly diviible by g (x)

## If $2x^{3}+ax^{2}+bx-6$ has (x-1) as a factor and leaves a remainder 2 when divided by x-2 find the relation between a and b.

Let, $f(x) = 2x^{3}+ax^{2}+bx-6$

$\because (x-1)$ is a factor of  f (x)  then by factor theorem f(1) = 0

f(1)=0

On putting  x = 1 in (1) we get

$f (1)=2+a+b-6$

$f (1)=a+b-4$

From (ii)   $f (1)=0\rightarrow a+b -4 = 0$

When f(x) is divided by (x-2)it leaves remainder 2

$\therefore$ by remainder theorem f(2) = 2

On putting x = 2 in (i)  and using (4) we get

$f(2)= 2(2)^{3}+a(2)^{2}+b(2)-6 = 2$

$\rightarrow 16+4a +2b-6 = 2$

$\rightarrow 4a+2b+8 = 0$

$\rightarrow 2a+b+4 = 0$

From (ii) and (v)

$2a+b+a+b = 0$

$\rightarrow 3a+2b = 0$

$\rightarrow 3a = -2b$

## If  x+a is a factor of the polynomial $x^{2}+px+q$ and $x^{2} +mx+n,$ prove that $a =\frac{n-q}{m-p}$

Let,

$p (x)= x^{2}+px+q$

and $q (x)= x^{2}+mx+n$

Since x+a is a factor  of p(x) and q(x) then p(-a) = 0 and q (-a) = 0

$p(-a) = 0\rightarrow (-a)^{2}+p (-a)+q=0$

$a^{2}-pa +q = 0$

$q(-a)=0 \rightarrow (-a)^{2}+m (-a)+n=0$

$a^{2}-ma+n = 0$

Subtracting eqn (2) from eqn (1) we get

$a^{2}-pa-a^{2}+ma-n = 0$

$\rightarrow a (m-p) = n-q$

$a = \frac{n-q}{m-p}$

## Using appropriate identity factorize $4x^{2}-\frac{y^{2}}{9}$

$4x^{2}-\frac{y^{2}}{9} = (2x)^{2}-\left ( \frac{y}{3} \right )^{2}$

$= \left ( 2x+\frac{y}{3} \right ) \left ( 2x-\frac{y}{3} \right )$

## Simplify : $\left ( x+\frac{1}{2} \right )\left ( x+\frac{3}{2} \right )$ .

$\left ( x+\frac{1}{2} \right )\left ( x+\frac{3}{2} \right ) = x^{2}+\frac{3}{2} x +\frac{1}{2}x+\frac{3}{4}$

$= x^{2}+2x+\frac{3}{4}$

## Write the co efficient of  $x^{2}$ in the expression of  $(x-2)^{3}$

$(x-2)^{3} = (x)^{3}-(2)^{3}-3\times x\times 2(x-2)$

$= x^{3}-8-6x^{2}+12x$

coefficient of $x^{2}$in the expansion of  $x- 2 ^{3} =-6$

## If  (a+b +c) = 0 then write equivalent  of $a^{3}+b^{3}+c^{3}.$

a$a^{3}+b^{3}+c^{3}-3abc$

$= (a+b+c) (a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$a^{3}+b^{3}+c^{3}-3abc = 0, a+b+c = 0$

$a^{3}+b^{3}+c^{3} = 3abc$

## If  $x+\frac{1}{x} = 4$ then calculate the value  of $x^{2}\frac{1}{x^{2}}$

$x^{2}+\frac{1}{x^{2}} = \left ( x+\frac{1}{x} \right )^{2}-2(x) \left ( \frac{1}{x} \right )$

$= (4)^{2}-2 = 16-2 = 14$

## If  $\frac{x}{y}+\frac{y}{x} = -1$ $(x\neq y, y \neq 0, x\neq 0)$ then  what is the value of $x^{3}-y^{3}$

Given,

$\frac{x}{y}+\frac{y}{x} = -1$

$\rightarrow \frac{x^{2}+y^{2}}{yx} = -1$

$\rightarrow x^{2}+y^{2}= -xy$

$\rightarrow x^{2}+y^{2} +xy = 0$

$\rightarrow x^{3}-y^{3} +xy = (x-y) (x^{2}+y^{2}+xy )$

$(x-y)$x$0 = 0$

## Calculate the value of  $\frac{83^{3}+17^{3}}{83^{2}-83\times 17+17^{2}}$

$\frac{83^{3}+17^{3}}{83^{2}-83\times 17+17^{2}}$=$\frac{(83+17) (83^{2}-83 \times 17+17^{2} )}{(83^{2}-83\times 17+17^{2})}$

$\because a^{3}+b^{3} = (a+b) (a^{2}+b^{2}-ab)$

$83+17 = 100.$

## Find the value of polynomial $x^{2}-9$ for x = 97.

$x^{2}-9 = (97)^{2}-(3)^{2}$

$= (97+3) (97-3)$

$= 100\times 94$

$= 9400$

## Evaluate  $249\times 251$ by using identity

$249\times 251=(250-1)(250+1)$

$=(250)^{2}-(1)^{2}$

$=62500-1$

$=62499.$

## Expand by using identity (2x-y+z)2

$(2x-y+z)^{2}=4x^{2}+y^{2}+z^{2}-4xy+2yz+4zx$

Alternative method

By using this identity

$(a+b+c)^{2}$

$a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

$(2x+(-y)+z)^{2} = (2x)^{2}+(-y)^{2}+c^{2}+2(2x)(-y)+2(-y)(z) + 2(z)(2x)$

$=4x^{2}+y^{2}+2xy$

## Expand using suitable identity (2x-3y+z)2.

$(2x-3y+z)^{2}=[2x+(-3y)+z]^{2}$

$=(2x)^{2}+(3y)^{2}+z^{2}+2\times 2x \times (-3y)+2\times (-3y)\times z+2\times 2x\times z$

$=4x^{2}+9y^{2}+z^{2}-12xy-6yz+4xz$

## Expand $\left ( \frac{a}{4} - \frac{b}{2}+1\right )$ Using identity

$\left ( \frac{a}{4} - \frac{b}{2}+1\right )$$\left ( \frac{a}{4} - \frac{b}{2}+1\right )^{2}=[\frac{a}{4}+\left ( -\frac{b}{2}+1 \right )]^{2}$

$=\left ( \frac{a}{4} \right )^{2}+\left ( -\frac{b}{2} \right )^{2}+(1)^{2}+2\times \frac{a}{4}\times \left ( -\frac{h}{2} \right )+2\times \left ( \frac{-b}{2} \right ) \times (1)+2 \times$$\left ( \frac{a}{4} \right )\times (1)$

$=\frac{a^{2}}{16}+\frac{b^{2}}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$

## Give possible expression for the length and breadth  of a rectangle whose  area is given by  $25a^{2}-35a+12$

Area of rectangle = $25a^{2}-35a+12$

=$25a^{2}-20a-15a+12$

$=5a (5a-4)-3(5a-4)$

$= (5a-3) (5a-4)$

$= length \times breadth$

$\therefore$ length and bredth are (5a-3) and (5a-4) respectively .

## Expand $\left ( \frac{1}{3}x- \frac{2}{3}y \right )^{3}$

$\left ( \frac{1}{3}x-\frac{2}{3}y \right )3$

$\left ( \frac{1}{3}x \right )^{3}-\left ( \frac{2}{3}y \right )^{3}-3\times \frac{1}{3}x\times \frac{2}{3y} \left ( \frac{1}{3}x-\frac{2y}{3} \right )$

$=\frac{x^{3}}{27}-\frac{8y^{3}}{27}- \frac{2xy}{3}\left ( \frac{x}{3}-\frac{2y}{3} \right )$

$=\frac{x^{3}}{27}-\frac{8y^{3}}{27}- \frac{2x^{2}y}{9}+\frac{4xy^{2}}{9}$

## using suitable identity evaluate (103)3

$103^{3}= (100+3)^{3}$

$103^{3}=3^{3}+3\times 100\times 3(100+3)$

$1000000+27+900\times 103$

$1000000+27+92700$

$1092727.$

## If  x and y are two positive real numbers such that x2+4y2 = 17 and xy =2then find the value of (x+2y)

$(x+2y)^{2}=x^{2}+4y^{2}+2\times x\times 2y$

$=17+4\times 2$

$=17+8=25$

$\Rightarrow (x+2y) = \pm \sqrt{25}=\pm 5.$

## Simplify $\left ( x+\frac{1}{x} \right )\left ( x-\frac{1}{x} \right )\left ( x^{2}+\frac{1}{x^{2}} \right )\left ( x^{4}+\frac{1}{x^{4}} \right )$

$\left ( x+\frac{1}{x} \right )\left ( x-\frac{1}{x} \right )\left ( x^{2}+\frac{1}{x^{2}} \right )\left ( x^{4}+\frac{1}{x^{4}} \right )$

$=\left ( x^{2}-\frac{1}{x^{2}} \right )\left (x^{2}+\frac{1}{x^{2}} \right )\left ( x^{4}+\frac{1}{x^{4}} \right )$

as (a+b)(a-b)= a2-b2

$=\left ( x^{4}-\frac{1}{x^{4}} \right ) \left ( x^{4}+\frac{1}{x^{4}} \right )$

$\left ( x^{8}+\frac{1}{x^{8}} \right )$

## Find the value of $x^{2}+\frac{1}{x^{2}},$ if $x-\frac{1}{x} = \sqrt{3}$

Given,

$x-\frac{1}{x} = \sqrt{3}$

On squaring both sides, we get

$\left ( x-\frac{1}{x} \right )^{2} = (\sqrt{3})^{2}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}-2\times x\times \frac{1}{x}= 3$

$\Rightarrow x^{2}+\frac{1}{x^{2}}= 3+2 =5.$

## if a,b,c, are all non-zero and a+b+c = 0 prove  $\frac{a^{2}}{bc}+\frac{b^{2}}{ac}+\frac{c^{2}}{ab} = 3$

$\frac{a^{2}}{bc}+\frac{b^{2}}{ac}+\frac{c^{2}}{ab} = 3$

$\frac{a^{2}}{bc}+\frac{b^{2}}{ac}+\frac{c^{2}}{ab} = \frac{a^{3}+b^{3}+c^{3}}{abc}$

$=\frac{3abc}{abc}$

$\left ( \because a+b+c = 0 \right )$

$\left ( \because a^{3}+b^{3}+c^{3} = 3abc \right )$

=3

## Factorize : 12(x2+7)2 - 8 (x2+7) (2x-1)-15 (2x-1)2.

Let $x^{2}+7=p and 2x-1 = q$ the given expression

$=12p^{2}-8pq-15q^{2}$

$=12p^{2}-18pq+10pq-15q^{2}$

$=6p(2p-3q) 5q(2p-3q)$

$=(2p-3q)(6p+5q)$

$= [2(x^{2}+7)-3(2x+1)] [6(x^{2}+7)+5 (2x-1)]$

$= (2x^{2}+14-6x+3) (6x^{2}+42+10x-5+1)$

$= (2x^{2}-6x+17 ) (6x^{2}+10x+37)$

## Without actually calculating the cubes . Evaluate $14^{3}+13^{3}-27^{3}$ .

$14=a, 13=b, -27=c$.

$a+b+c = 14+13-27 = 0$

$a^{3}+b^{3}+c^{3} = 3abc$$14^{3}+13^{3}-27^{3}= 3\times 14\times 13\times (-27)$

$= -14742.$

## If a = 3+b then  that is the value of $a^{3}-b^{3}-9ab$

$a = 3 +b$

$\rightarrow a -b = 3$

$\rightarrow (a -b)^{3} = (3)^{3}$

$a^{3}-b^{3}-3ab(a-b)= 27$

$a^{3}-b^{3}-3ab \times 3 = 27$

$a^{3}-b^{3}-9ab = 27$

## Factorize : m (m-1) -n (n-1)

$m (m-1) -n (n-1)=m^{2}-m-n^{2}+n$

$=m^{2}-n^{2}-m+n$

$(m-n)(m+n)-(m-n)$

$=(m-n)(m+n-1).$

## factorize : $9x^{2}+6xy+y^{2}$

$9x^{2}+6xy+y^{2} = (3x)^{2}+2\times (3x)\times y+y^{2}$

$=(3x+y)^{2}$

$\left [ \because a^{2}+2ab +b^{2}=(a+b)^{2}\right ]$

## Factorize : $8x^{3}-(2x-y)^{3}$

$8x^{3}-(2x-y)^{3}=(2x)^{3}-(2x-y)^{3}$

$= [2x-(2x-y)][(2x)^{2}+(2x-y)^{2}+2x(2x-y)]$

$(a^{3}-b^{3})= (a-b) (a^{2}+b^{2}+ab)$

$=y [4x^{2} +4x^{2}+y^{2}-4xy+4x^{2}-2xy]$

$=y [12x^{2}+y^{2}-6xy]$

## Factorize  $x^{4}-y^{4}$

$x^{4}-y^{4}=(x^{2})^{2}-(y^{2})^{2}$

$= (x^{2}-y^{2}) (x^{2}+y^{2})$

$= (x-y)(x+y) (x^{2}+y^{2})$

## factorize $64x^{3}+\sqrt{125y^{3}}$

$64x^{3} + \sqrt{125} y^{3}$

$=(4x+ \sqrt{5}y) (16x^{2}-4 \sqrt{5}xy+5y^{2})$

Alternative method

$64x^{3}+ \sqrt{124}y^{3} =(4x)^{3} + ( \sqrt{5}y)^{3}$

$= (4x+ \sqrt{5}y)[(4x)^{2}+( \sqrt{5}y)^{2}-4x$ x $\sqrt{5}y$

$= (4x+ \sqrt{5}y)[16x^{2}+5y^{2}-4 \sqrt{5}xy]$.

## factorize $x^{4}-125xy3$

$x^{4}-125xy^{3}= x(x^{3}-125y^{3})$

$x [(x)^{3}-(5y)^{3}]$

$x (x-5y)[x^{2}+(5y)^{2}+x \times 5y]$

$x (x-5y)[x^{2}+(25y)^{2}+ 5xy]$

## Factorize $2y^{3}+y^{2}-2y-1$

$2y^{3}+y^{2}-2y-1 = 2y^{3}-2+y^{2}-2y+1$

$=2 (y-1) (y^{2}+y+1)+(y-1)^{2}$

$= (y-1) [2y^{2}+3y+1]$

$= (y-1) (y+1) (2y+1)$

## Factorize 8-27a3-36a+54a2

8-27a3-36a+54a2

$=(2)^{3}-(3a)^{3}-18a (2-3a)$

$=(2)^{3}-(3a)^{3}-3\times 2\times 3a (2-3a)$

$=(2-3a)^{3}$

## Factorize  64a3-27b3-144a2b+108ab2

64a3-27b3-144a2b+108ab2

$= (4a)^{3}-(3b)^{3}-3\times (4a)^{2}\times (3b)+3\times (4a)\times (3b)^{2}$

$= (4a)^{3}-(3b)^{3}-3\times 4a\times 3b(4a-3b)$

$=(4a-3b)^{3}$

## Factorize 3-12(a-b)2

3-12(a-b)$= 3[1-4 (a-b)^{2}]$

$=3[(1)^{2}-\left \{ 2(a-b) \right \}^{2}]$

$=3[1+ 2(a-b )] [1-2 (a-b)]$

## Factorize  a6-b6

a$a^{6}-b^{6} = (a^{3})^{2}-(b^{3})^{2}$

$= (a^{3}-b^{3}) (a^{3}+b^{3})$

$= (a-b) (a^{2}+b^{2}+ab) (a+b) (a^{2}+b^{2}-ab)$

$= (a-b) (a+b) (a^{2}+b^{2}+ab)(a^{2}+b^{2}-ab)$

## Factorize 250x3-432y3.

$250x^{3} - 432y^{3}=2[125x^{3}-216y^{3}]$

$2[(5x)^{3}-(6y)^{3}]$

$2[(5x-6y) [(5x)^{2}+(6y)^{2}+5x \times 6y]$

$\because a^{3}-b^{3}=(a-b) (a^{2}+b^{2}+ab)$

$=2(5x-6y)[25x^{2}+36y^{2}+30xy]$

## Factorize $p^{3}q^{3} +\frac{343}{729}=(pq)^{3}+\left ( \frac{7}{9} \right )^{3}$

$p^{3}q^{3} +\frac{343}{729}=(pq)^{3}+\left ( \frac{7}{9} \right )^{3}$

$=\left ( pq + \frac{7}{9} \right )\left ( (pq)^{2}+\left ( \frac{7}{9} \right )^{2}-pq\times \frac{7}{9} \right )$

$\because a^{3}+b^{3} = (a+b) (a^{2}+b^{2}-ab)$

$=\left ( pq+\frac{7}{9} \right )\left ( p^{2}q^{2}+\frac{49}{81}-\frac{7pq}{9} \right )$

## Factorize  (x-y)2-7(x2-y2)+12(x+y)2

Given Exp = $(x-y)^{2}-7(x+y)(x-y)+12(x+y)^{2}$

$=(x-y) -4(x+y)(x-y)-3(x+y)(x-y)+12(x+y)^{2}$

$=(x-y) [x-y-4x-4y]-3(x+y)[x-y-4x-4y]$

$=(x-y)[-5y-3x]-3(x+y)[-3x-5y]$

$[-5y-3x]x-y-3(x+y)$

$=(-5y-3x) (-2x-4y)$

$=(5y+3x) (2x+4y)$

$=2(x+2y) (5y+3x)$

## If x+y+z = 0 Show that x3+y3+z3 = 3xyz

$x+y+z = 0$

x+y = -z

$(x+y)^{3}= (-z)^{3}$

$x^{3}+y^{3} +3xy (-z) = -z^{3}$

$x^{3}+y^{3} -3xyz = -z^{3}$

$x^{3}+y^{3} +z^{3} = 3xyz$

## Factorize (i) $x^{2}+\frac{1}{x^{2}} +2-2x- \frac{2}{x}$  (ii) $x^{4}-y^{4}$

(i)  $x^{2}+\frac{1}{x^{2}}+2-2x-\frac{2}{x}$

$(x)^{2}+\left ( \frac{1}{x} \right )^{2}+2\times x\times \frac{1}{x}-2x-\frac{2}{x}$

$= \left ( x+\frac{1}{x} \right )^{2}-2 \left ( x+ \frac{1}{x}\right )$

$= \left ( x+\frac{1}{x} \right ) \left ( x+ \frac{1}{x}-2\right )$

(ii)  $x^{4}-y^{4} = (x^{2})^{2}-(y^{2})^{2}$

$(x^{2}-y^{2}) (x^{2}+y^{2})$

$= (x-y) (x+y) (x^{2}+y^{2})$

## Factorize  $(x^{2}-4x) (x^{2}-4x-1)-20$

Let, $x^{2}-4x= a$

$\therefore$ expression $= a (a-1)-20$

$=a^{2}-a-20$

$=(a-5)(a+4)$

$= (x^{2}-4x-5) (x^{2}-4x+4)$

$= (x-5) (x+1) (x-2)^{2}$

Alternative method

$(x^{2}-4x) (x^{2}-4x-1)-20 = (x^{2}-4x )^{2}-(x^{2}-4x)-20$

$= (x^{2}-4x)^{2}-5(x^{2}-4x)+4 (x^{2}-4x)-20$

$= (x^{2}-4x) [x^{2}-4x-5]+4 [x^{2}-4x-5]$

$= (x^{2}-4x-5) (x^{2}-4x+4)$

$= (x^{2}-5x+x-5)[(x)^{2}-2\times 2\times x+(2)^{2}]$

$= [x(x-5)+1(x-5)] [x-2]^{2}$

$= [(x-5)(x+1)(x-2)(x-2)$

## Factorize : $9x^{2}+y^{2}+z^{2}-6xy+2yz-6xz.$ Hence find its value when x = 1, y= 2 and z  = -1

$9x^{2}+y^{2}+z^{2}-6xy+2yz-6xz.$

$=(-3x)^{2}+(y)^{2}+(z)^{2}+2\times (-3x) (y)+2\times (y)(z)+2(-3x)(z)$

$= (-3x+y+z)^{2}$

If  $x = 1, y= 2, z = -1$ then

$(-3x+y+z)^{2}=(-3\times 1+2-1)^{2}$

$=(-3 +2-1)^{2}$

$=4$

## Factorize $27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p$

$27p^{3}-\frac{1}{216}-\frac{9}{2}p^{2}+\frac{1}{4}p$

$=(3p)^{3}-\left ( \frac{1}{6} \right )^{3}-3. (3p)^{2}\frac{1}{6}+3(3p)\left ( \frac{1}{6} \right )^{2}$

$=\left ( 3p-\frac{1}{6} \right )^{3}$

$=\left ( 3p-\frac{1}{6} \right )\left ( 3p-\frac{1}{6} \right )\left ( 3p-\frac{1}{6} \right )$

## Factorize 125x3-27y3+z3+45xyz

$125x^{3}-27y^{3}+z^{3}+45xyz$

$= (5x)^{3}+(-3y)^{3}+(z)^{3}-3\times (5x)(-3y) (z)$

$= (5x-3y+z) [(5x)^{2}+(-3y)^{2}+(z)^{2}-(5x)(-3y)-(-3y)(z)-(5x)(z)]$

$\because a^{3}+b^{3}+c^{3}-3abc$

$=(a+b+c) (a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$= (5x-3y+z)[25x^{2}+9y^{2}+z^{2}+15xy+3yz-5xz]$

## if x+y+z = 0, show that x3+y3+z3 =3xyz

$x+y+z = 0$

$\Rightarrow x+ y = -z$

$\Rightarrow (x+y)^{3} = (-z)^{3}$

$\Rightarrow x^{3}+y^{3}+3xy(-z) =-z^{3}$

$\Rightarrow x^{3}+y^{3}-3xyz =-z^{3}$

$\Rightarrow x^{3}+y^{3}-z^{3} = 3xyz$

## Using a suitable identity, find (98)3

$(98)^{3}= (100-2)^{3}$

$= (100)^{3}-(2)^{3}-3\times 10\times 2 (100-2)$

$= 1000000-8-600\times 98$

$= 1000000-8-58800$

$= 1000000-58808=941192$

## Evaluate  1113, using a suitable identity

$111^{3}= (100+11)^{3}$

$(100)^{3}+3(100)^{2}(11)+3(100)(11)^{2}+(11)^{3}$

$=1367631$

## Evaluate : $(\sqrt{2}+\sqrt{3})^{2}+(\sqrt{5}-\sqrt{2})^{2}$

$(\sqrt{2}+\sqrt{3})^{2}+(\sqrt{5}-\sqrt{2})^{2}$

$=(\sqrt{2})^{2}+(\sqrt{3})^{2}+2\times \sqrt{2}\times \sqrt{3}+(\sqrt{5})^{2}+(\sqrt{2})^{2}-2\times \sqrt{5}\times \sqrt{2}$

$=2+3+2\sqrt{6}+5+2-2\sqrt{10}$

$=12+2\sqrt{6}-2\sqrt{10}$

$=2(6+\sqrt{6}-\sqrt{10})$

## Simplify  :(2a+3b)3-(2a-3b)3

$(2a+3b)^{3}-(2a-3b)^{3}=x^{3}-^{3}$ where  2a + 3b = x and 2a -3b = y

$(x-y) (x^{2}+xy+y^{2})$

$[(2a+3b)-(2a-3b)] [(2a+3b)^{2}+(2a+3b) (2a-3b)+(2a-3b)^{2}]$

$= 6b[(4a^{2}+12ab+9b^{2})+(4a^{2}-9b^{2})+(4a^{2}-12ab+9b^{2})]$

$= 6b(12a^{2}+9b^{2})$

$= 6b\times 3\times (4a^{2}+3b^{2})$

$= 18b(4a^{2}+3b^{2})$

## If $x^{2}+\frac{1}{x^{2}}= 7,$ Find the value of $x^{3}+\frac{1}{x^{3}}$ taking only the positive value of  $x+\frac{1}{x}$

$\left ( x+\frac{1}{x} \right )^{2}=x^{2}+\frac{1}{x^{2}}+2x\frac{1}{x}$

$=x^{2}+\frac{1}{x^{2}}+2$

$=7+2 =9$

$\left ( x+\frac{1}{x} \right ) = \pm 3$

$( x+\frac{1}{x} =3$ [On taking positive value ]

Now,  $x^{3}+\frac{1}{x^{3}} = \left ( x+\frac{1}{x} \right )\left ( x^{2}+\frac{1}{x^{2}}-x\frac{1}{x} \right )$

$=\left ( x+ \frac{1}{x} \right ) \left ( x^{2}+\frac{1}{x^{2}}-1 \right )$

$=(3)(7-1)$

= 3 x 6 = 18

## If  $x^{2}+\frac{1}{x^{2}}=98$ , then find the value of $x^{3}+\frac{1}{x^{3}}$

$x^{2}+\frac{1}{x^{2}} = 98$

$\rightarrow \left ( x+\frac{1}{x} \right )^{2} = 98+2 = 100$

$x+\frac{1}{x} = 100$

$x^{3}+\frac{1}{x^{3}} = \left ( x+\frac{1}{x} \right )\left ( x^{2}-x\frac{1}{x}+\frac{1}{x^{2}} \right )$

$=(10)\left ( x^{2}+\frac{1}{x^{2}}-1 \right )$

$=(10)(98-1)$

= 10 x 97 = 970

## If z2+$z^{2}+\frac{1}{z^{2}}=14$ Find the value $z^{3}+\frac{1}{z^{3}}$  taking positive value of $z+\frac{1}{z}$

$\left ( z+\frac{1}{z} \right )^{2}= z^{2}+\frac{1}{z^{2}}+2$

$=14+2 =16$

$z^{3}+\frac{1}{z^{3}}= \left ( z+\frac{1}{z} \right )^{3}-3 \left (z+ \frac{1}{z} \right )$

$=(4)^{3}-3(4)$

$=64-12$

$\rightarrow z^{3}+\frac{1}{z^{3}} = 52$

## Simplify  : $\left ( \frac{x}{3}+\frac{y}{5} \right )^{3}-\left ( \frac{x}{3}-\frac{y}{5} \right )^{3}$

$\left ( \frac{x}{3}+\frac{y}{5} \right )^{3}-\left ( \frac{x}{3}-\frac{y}{5} \right )^{3}$$= [\frac{x}{3}+\frac{y}{5}-\frac{x}{3}+\frac{y}{5}]$

$\left [ \left ( \frac{x}{3}+\frac{y}{5} \right )^{2}+\left ( \frac{x}{3}-\frac{y}{5} \right )^{2}+\left ( \frac{x}{3}+\frac{y}{5} \right ) \left ( \frac{x}{3}-\frac{y}{5} \right )\right ]$

$=\left ( \frac{2y}{5} \right )\left [ \frac{x^{2}}{9}+\frac{y^{2}}{25}+2\times \frac{x}{3}\times \frac{y}{5} +\frac{x^{2}}{9}+\frac{y^{2}}{25}-2\times \frac{x}{3}\times \frac{y}{5}+\frac{x^{2}}{9}-\frac{y^{2}}{25}\right ]$

$=\frac{2y}{5}\left [ 3\times \frac{x^{2}}{9}+\frac{y^{2}}{25} \right ]=\frac{2y}{5}\left ( \frac{x^{2}}{3}+\frac{y^{2}}{25} \right )$

## If $a^{2}+b^{2}+c^{2}= 280$ and ab+bc+ca = $\frac{9}{2}$ then find the value of $(a+b+c)^{3}$

$(a+b+c)^{2} =$$a^{2}+b^{2}+c^{2}$$+2(ab+bc+ca) = 280++2 \frac{9}{2}$

$\Rightarrow (a+b+c)^{2} = 280+9 = 289$

$\Rightarrow a+b+c = \sqrt{289}= 17$

$\Rightarrow (a+b+c )^{3} = 17^{3} = 4913$

## Simplify : (3a-2b) (9a2+6ab+4b2)-(2a+3b) (4a2-6ab+9b2).

$(3a-2b)(9a^{2}+6ab+4b^{2})-(2a+3b) (4a^{2}-6ab+ab^{2})$

$\Rightarrow \left [ (3a)^{3}-(2b)^{3} \right ]-\left [ (2a)^{3}+(3b)^{3} \right ]$

$\Rightarrow \left [ 27a^{3}-8b^{3} \right ]-\left [ 8a^{3}+27b^{3} \right ]$

$\Rightarrow 27a^{3}-8b^{3}-8a^{3}-27b^{3}$

$\Rightarrow 19a^{3}- 35b^{3}$

## If x+ y+4 = 0, then find the value of $x^{3}+y^{3}-12xy+64.$

$x+y = -4$

$\Rightarrow x+y+ 4 = 0.$

$a+b+c = 0 \Rightarrow a^{3}+b^{3}+c^{3} = 3abc$

$\Rightarrow x^{3}+y^{3}+4^{3} = 3xy (4)$

i,e., $x^{3}+y^{3}-12xy +64 = 0$

Alternative method

$x^{3}+y^{3}-12xy +64 = ( x)^{3}+(y)^{3}+(4)^{3}-3 \times (x)(y) (4)$

$= (x+y+4) (x^{2}+y^{2}+16-xy-4y-4x)$

$= 0 \times (x^{2}+y^{2}+16-xy-4y-4x)=0$

## Find the value of $x^{3}+y^{3}+15xy-125,$ when x+y = 5.

Given, x+y = 5 then

$x^{3}+y^{3}+15xy-125$

$=x^{3}+y^{3}-125+15xy$

$=x^{3}+y^{3}+(-5)^{3}-3\times x\times y\times (-5)$

$= (x+y-5)\left [ (x)^{2}+(y)^{2}+(-5)^{2} \right -x\times (-5)-y\times (-5)-x\times y]$

$=(5-5)(x^{2}+y^{2}+25+5x+5y)$

$=0(x^{2}+y^{2}+25+5x+5y-xy)$

$=0$

## Simplify  $\frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}$

$\frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}$

Both numerator and denominator are of the form a3+b3+c3

We  know that when a+b+c = 0

then a3+b3+c3 = 3abc

For numerator, a2-b2+b2-c2+c2-a2= 0

For denominator     a-b+b-c+c-a= 0

$\therefore \frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}$

$=\frac{3\times (a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})}{3(a-b) (b-a)(b-c)}$

$=\frac{ (a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{(a-b) (b-c)(c-a)}$

$= (a+b) (b+c) (c+a)$

## If $x= \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$  and $y= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ find the value of $x^{2}-y^{2}+xy$ if $\sqrt{6}= 24$ .

$x^{2}-y^{2}+xy$   $\left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )^{2}-\left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )^{2}$

$+\left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right ) \times \left ( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \right )$

$=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}-\frac{5-2\sqrt{6}}{5+2\sqrt{6}}+1$

times

$=\frac{(5+2\sqrt{6})^{2}-(5-2\sqrt{6})^{2}}{(5-2\sqrt{6})(5+2\sqrt{6})} +1$

$=\frac{25+24+20\sqrt{6}-25-24+20\sqrt{6}}{25-24}+1$

$=40\sqrt{6}+1$

$=40\times 24+1=96+1=97$

## Simplify : $(a+2b+3c)^{2}-(a-2b-3c)^{2}-6b^{2}-9bc.$

$(a+2b+3c)^{2}-(a-2b-3c)^{2}-6b^{2}-9bc.$

$(a+2b+3c +a-2b-3c) (a+2b+3c-a+2b+3c)-6b^{2}-9bc$

$= 2a(4b+6c)-6b^{2}-9bc$

= 2a x 2 (2b+3c)-3b(2b+3c)

= (2b+3c) (4a-3b).

## Find the value ab + bc +ca, if  a + b + c = 9  and a2+b2+c2 = 35.

Given,

a + b + c = 9

$\Rightarrow (a+b+c)^{2} = 9^{2}$

$\Rightarrow a^{2}+b^{2}+c^{2}+2(ab+bc+ca) = 81$

$\Rightarrow as \left [ (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+ 2(ab+bc+ca) \right ]$

$\Rightarrow 35+2(ab+bc+ca) = 81$

$\Rightarrow 2(ab+bc+ca) = 81-35=46$

$\Rightarrow ab+bc+ca=\frac{46}{2}$

$\Rightarrow ab+bc+ca=23.$

## Find the product : $(x+y+2z) (x^{2}+y^{2}+z^{2}-xy-2yz-2zx).$

$(x+y+2z) (x^{2}+y^{2}+z^{2}-xy-2yz-2zx)=$$(x+y+2z) \left [ x^{2}+y^{2}+(2z)^{2}-x\times y-y\times 2z-x\times 2z \right ]$

We know that

$(a+b+c) (a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$=a^{2}+b^{2}+c^{2} =3abc$

$\therefore (x+y+2z) (x^{2}+y^{2}+(2z)^{2}-x\times y-y\times 2z-x\times 2z)$

$= (x)^{3}+(y)^{3}+(2z)^{3}-3\times x\times y\times 2z$

$= x^{3}+y^{3}+8z^{3}-6xyz$

## Find the value of (x-a)3+(x-b)3+(x-c)3-3(x-a) (x-b) (x-c), if a+b+c = 3x.

Given, a+b+c = 3x

or  3x-a-b-c = 0

$\therefore (x-a)^{3}+(x-b)^{3}+(x-c)^{3}-3(x-a)(x-b) (x-c)$

$[x-a+x-b+x-c] [(x-a)^{2}+(x-b)^{2}+(x-c)^{2}-(x-a)(x-b)-(x-b)$

$(x-c)-(x-a)(x-c)]$

$=[3x-a-b-c] [(x-a)^{2}+(x-b)^{2}+(x-c)^{2}-(x-a) (x-b)-(x-b)(x-c)-(x-a)(x-c)]$

$= 0\times [(x-a)^{2}+(x-b^{2})+(x-c)^{2}-(x-a)(x-b)-(x-b) (x-c)-(x-c)(x-a)]$

=0.

## Evaluate by using identities  (i)   103 x 107 (ii) (102)3

(i)$103 \times 107 = (100+3)(100+7)$

$100^{2}+(3+7)\times 100+3\times 7$

$= 10000+1000+21$

$= 11021$

(ii) $(102)^{3}=(100+2)^{3}$

$(100)^{3}+2^{3}+3\times 100\times 2(100+2)$

$=1000000+8+600+102$

$=1000000+8+61200$

$=1061208$

## If $x+\frac{1}{x} = 5$evaluate   $x^{2}+\frac{1}{x^{2}}$

$x+\frac{1}{x} = 5$

On squaring both sides , we get

$\left (x+\frac{1}{x} \right )^{2}= 5^{2}$

$\rightarrow x^{2}+\left ( \frac{1}{x} \right )^{2}+2\times x+\frac{1}{x} = 25$

$[\because (a+b)^{2}=a^{2}+b^{2}+2ab)]$

$\Rightarrow x^{2}+ \frac{1}{x^{2}} +2= 25$

$\Rightarrow x^{2}+ \frac{1}{x^{2}} = 25-2$

$\Rightarrow x^{2}+ \frac{1}{x^{2}} =23$

## If $x+\frac{1}{x}= \sqrt{3},$ evaluate $x^{3}+\frac{1}{x^{3}} .$

$\left (x+\frac{1}{x} \right )^{3} =\sqrt{3}$

Cubing both sides

$\left (x+\frac{1}{x} \right )^{3} =\left (\sqrt{3} \right )^{3}$

$\Rightarrow (x^{3}+\frac{1}{x^{3}}+3x\frac{1}{x} \left ( x+\frac{1}{x} \right )=3\sqrt{3}$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+3\left ( \sqrt{3} \right )=3\sqrt{3}$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=0$

## If  $x -\frac{1}{x}=2,$ find $x^{4} -\frac{1}{x^{4}},$

$x^{2}+\frac{1}{x^{2}}=\left ( x-\frac{1}{x} \right )^{2}+2$

=4+2 = 6

$\left (x^{2}+\frac{1}{x^{2}} \right )^{2}=\left ( x^{4}-\frac{1}{x^{4}} \right )+2$

$\rightarrow (6)^{2}=x^{4}+\frac{1}{x^{4}}+2$

$\rightarrow 36-2=x^{4}+\frac{1}{x^{4}}$

$\rightarrow x^{4}+\frac{1}{x^{4}} =34$

## If $x^{3}+\frac{1}{x^{2}} = 7,$ find the value $x^{3}+\frac{1}{x^{3}} = 7,$

$x^{2}+\frac{1}{x^{2}} = 7,$

$x^{2}+\frac{1}{x^{2}} +2= 7+2 = 9$

$\rightarrow \left (x +\frac{1}{x } \right )^{2} = 9$

$\rightarrow x + \frac{1}{x} =\sqrt{9}=3$

and  $\left ( x+\frac{1}{x} \right )^{3} = x^{3}+\frac{1}{x^{3}}+3\left ( x+\frac{1}{x} \right )$

$\Rightarrow 3^{3}= x^{3}+\frac{1}{x^{3}}+3\times 3$

$\Rightarrow x^{3}= \frac{1}{x^{3}}=27-9=18$

## Prove that $(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}=3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)$

$a^{2}-b^{2}, y = b^{2}-c^{2}, z=c^{2}-a^{2}$

Now,  $x+y+z = a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2} = 0$

$\because x+y+z = 0$

$\Rightarrow x^{3}+y^{3}+z ^{3}= 3xyz$

$i,e., (a^{2}-b^{2} )^{3}+(b^{2}-c^{2}) ^{3}+(c^{2}-a^{2})^{3}$

$=3(a^{2}-b^{2} ) (b^{2}-c^{2}) (c^{2}-a^{2})$

$=3(a +b ) (a-b ) (b+c)(b-c) (c+a)(c-a)$

$=3(a +b ) (b+c ) (c+a)(a-b) (b-c)(c-a)$

## Prove that $2x^{3}+2y^{3}+2z^{3}-6xyz=(x+y+z)$ $[(x-y)^{2 } +(y-z)^{2}+(z-x)^{2}].$ Hence evaluate : $2(13)^{3}+2(14)^{3}+2(15)^{3}-6\times 13\times 14\times 15$

Prove RHS =LHS as in Q.2

Now, $2(13)^{3}+2(14)^{3}+2(15)^{3}-6\times 13\times 14\times 15$

$=(13+14+15)[(13-14)^{2}+(14-15)^{2}+(15-13)^{2}]$

=42 x [1+1+4]

= 42 x 6 = 252.

## Simplify (a+b)3+(a-b)3+6a(a2-b2).

$(a+b)^{3.}+(a-b)^{3}+6a(a^{2}-b^{2})$

$(a+b)^{3.}+(a-b)^{3}+3\times 2a(a-b)(a+b)$

$=(a+b)^{3.}+(a-b)^{3}+3(a+b)(a-b) \left [ (a+b)+(a-b) \right ]$

$=[(a+b)+(a-b)]^{3}$

$=(2a)^{3}$

$=8a^{3}$

## Find the value of $x^{3}-8y^{3}-36xy-216$when $x = 2y+6.$

$x^{3}-8y^{3}-36xy-216$

$=(x)^{3}+(-2y)^{3}+(-6)^{3}-3(x)(-2y)(-6)$

$=[x+(-2y)+(-6)][x^{2}+(-2y)^{2}+(-6)^{2}-(x)(-2y)-(-2y)(-6)-(x)(-6)]$

$=(x-2y-6)(x^{2}+4y^{2}+36+2xy-12y+6x)$

=0 x $[x^{2}+4y^{2}+36+2xy-12y+6x]$

$\left (\because x = 2xy+6 or x-2y-6=0 \right )$

$\therefore x^{3}-8y^{3}-36xy-216=0.$

## Prove that $x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z)$ $\left [ (x-y)^{2}+(y-z)^{2}+(z-x)^{2} \right ]$

RHS = $\frac{1}{2}(x+y+z)[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]$

$=\frac{1}{2}(x+y+z) [x^{2}+y^{2}-2xy+y^{2}+z^{2}-2yz+z^{2}+x^{2}-2zx]$

$=\frac{1}{2}(x+y+z) [2x^{2}+2y^{2}+2z^{2}-2xy -2yz+2zx]2$

$=\frac{1}{2}(x+y+z)2 [x^{2}+y^{2}+z^{2}+xy-yz-zx]$

$=x^{3}+y^{3}+z^{3}-3xyz$

$= LHS$

## if a+b+c = 6 and ab+bc+ca = 11, find the value of a3+b3+c3-3abc.

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

$(6)^{2}=a^{2}+b^{2}+c^{2}+2\times 11$

$=a^{2}+b^{2}+c^{2}=36-22=14$

$=a^{3}+b^{3}+c^{3}-3abc = (a+b+c)[a^{2}+b^{2}+c^{2}-ab+bc+ca]$

$=6\times (14-11)=6\times 3=18.$

Alternative method

$a^{3}+b^{3}+c^{3}-3abc= (a+b+c) [a^{2}+b^{2}+c^{2}-(ab+bc+ca)]$

$=(a+b+c) [(a+b+c)^{2}-3(ab+bc+ca)]$

$=6[6^{2}-3(11)]6[36-33]=6\times 3=18.$

## Find the value of p3-q3 , if p-q = $\frac{10}{9}$ and $pq \frac{5}{3}$

$p-q=\frac{10}{9}$   $pq=\frac{5}{3}$                  $p^{3}q^{3}= ?$

$(p-q)^{3}=p^{3}-q^{3}-3-pq(p-q)$

$\rightarrow \left ( \frac{10}{9} \right )^{3} p^{3}-q^{3}-3\times \frac{5}{3}\times \frac{10}{9}$

=   $\frac{5050}{729}$

Alternative method

$p^{3}-q^{3}=(p-q)(p^{2}-q^{2}+pq)$

$= (p-q)( (p-q)^{2}+3pq)$

$= \frac{10}{9}\left [ \left ( \frac{10}{9} \right ) ^{2}+3\times \frac{5}{3}\right ]$

$= \frac{10}{9} \left [ \frac{100}{81}+5 \right ]$

$= \frac{10}{9} \left [ \frac{100+405}{81} \right ]$

$= \frac{5050}{729}$

## If x+y+z = 10 and $x^{2}+y^{2}+z^{2}=40.$ find xy+yz+zx and $x^{3}+y^{3}+z^{3}-3xyz.$

Expansion of  (x+y+z)2

= $= x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$

$\rightarrow 100=40+2(xy+yz+zx)$

$\rightarrow (xy+yz+zx) = 30$

$\rightarrow x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

= $10(40-30)= 100$

Alternative method

$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$

$\rightarrow (10)^{2}=40+2(xy+yz+zx)$

$\rightarrow 100-40 = 2 (xy+yz+zx)$

$\rightarrow (xy+yz+zx) = \frac{60}{2}= 30$

and $\rightarrow x^{3}+y^{3}+z^{3}-3xyz$

$=(xy+yz+zx) = x^{2}+y^{2}+z^{2}-(xy+yz+zx) ]$

$=10[40-30]$

10x 1 0 = 100

## If$a = 5+2\sqrt{6}$ and $b=\frac{1}{a}$ then what will be value of $a^{2}+b^{2}$ and $a^{3}+b^{3}$

$b= \frac{1}{a}= \frac{1}{(5+2\sqrt{6})}\times \frac{(5-2\sqrt{6})}{(5-2\sqrt{6})}$

$\frac{(5-2\sqrt{6})}{25-24}=5-2\sqrt{6}$

$\therefore a+b = 5+2\sqrt{6}+5-2\sqrt{6}=10$

Also $b =\frac{1}{a}\Rightarrow ab = 1$

$a^{2}+b^{2} = (a+b)^{2}-2ab$

$=(10)^{2}-2\times 1$

$=100-2=98$

$(a^{3}+b^{3}) = (a+b)^{3}-3ab (a+b)$

$(10)^{3}-3\times 1\times 10$

$1000-30=970$

## Prove that : $(x+y)^{3}+(y+z)^{3}+(z+x)^{3}-3(x+y)(y+z)(z+x)=2(x^{3}+y^{3}+z^{3}-3xyz)$

$(x+y)^{3}+(y+z)^{3}+(z+x)^{3}-3(x+y)(y+z)(z+x)$

Since,

$a^{3}+b^{3}+c^{3} -3abc$

$= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

$= (x+y+y+z+z+x) (x+y)^{2}+(y+z)^{2}+(z+x)^{2} [-(x+y)(y+z)-(y+z)(z+x)-(x+y)(z+x)]$

$2(x+y+z) [x^{2}+y^{2}+z^{2}+2xy+y^{2}+z^{2}+2yz+z^{2}+x^{2}+2zx-xy-xz-y^{2}-yz-yz-xy-z^{2}-zx-xz-x^{2}-yz-xy]$

$=2(x+y+z) [x^{2}+y^{2}+z^{2}-xy-yz-zx]$

$=2(x^{3}+y^{3}+z^{3} -3xyz.)$

## if x and y are two positive real numbers such that 8x3+27y3= 730 and 2x2y+3xy2=15, then evaluate 2x+3y.

Rightarrow

$(2x+3y)^{3} = (2x)^{3}+(3y) ^{3}+3\times 2x\times 3y(2x+3y)$

$8x^{3}+27y^{3}+18(2x^{2}y+3xy^{2})$

$=730+18\times 15$

$=730+270$

$\Rightarrow (2x+3y)^{3}=100$

$\therefore 2x+3y =\sqrt[3]{1000}=10$

## ab+bc+ca =0, find the value of  $\frac{1}{a^{2}-bc}+\frac{1}{b^{2}-ca} +\frac{1}{c^{2}-ac}$

$ab+bc+ca = 0$

$-bc=ab+ca$

$- ca=ab+bc$

$- ab =bc+ca$

and

now.

$\frac{1}{a^{2}-bc}+\frac{1}{b^{2}-ca} +\frac{1}{c^{2}-ab}$

$=\frac{1}{a^{2}+ab+ca}+\frac{1}{b^{2}+cb+bc} +\frac{1}{c^{2}+bc+ca}$

$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)} +\frac{1}{c(a+b+c)}$

$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)} +\frac{1}{c(a+b+c)}$

$=\frac{bc+ca+ab}{abc(a+b+c)}$

$=0$

## Verify x3-y3 = (x-y)(x2+y2+xy). Hence factorize 216x3-125y3

RHS = $(x-y)(x^{2}+y^{2}+xy)$

$x^{3}+xy^{2}+x^{2}y-x^{2}y-y^{3}-xy^{2}$

$= x^{3}-y^{3}= LHS$

Now, 216x3-125y$(6x)^{3}-(5y)^{3}$

$=(6x-5y)[(6x)^{2}+(5y)^{2}+6x\times 5y]$

$=(6x-5y) (36x^{2}+25y^{2}+30xy)$

## Factorize (i) 4a2-9b2-2a-3b (ii) a2+b2-2(ab-ac+bc).

(i) $4a^{2}-9b^{2}-2a-3b$

$(2a)^{2}-(3b)^{2}-(2a+3b)$

$= (2a-3b)(2a+3b)-(2a+3b)$

$= (2a+3b)(2a-3b-1)$

(ii) $a^{2}+b^{2}-2 (ab-ac+bc)$

$= a^{2}+b^{2}-2 ab+2ac2bc)$

$= (a-b)^{2}+2c(a-b)$

$= (a-b) + [(a-b)+2c]$

$= (a-b) (a-b+2c)$

## Factorize completely : x8-y8

$x^{8}-y^{8}=(x^{4})^{2}-(y^{4})^{2}$

$(x^{4}+y^{4})(x^{4}-y^{4})$

$(x^{4}+y^{4})[(x^{2})^{2}-(y^{2})^{2}]$

$(x^{4}+y^{4}) (x^{2}+y^{2})(x^{2}-y^{2})$

$(x^{4}+y^{4}) (x^{2}+y^{2}) (x+y) (x-y)$

## Factorize :(p+q)2-20(p+q)-125

Let, p+q = x then

Let $f(x)=(p+q)^{2}-20(p+q)-125$

$=x^{2}-20x-125$

$=x^{2}-25x+5x-125$

$= x(x-25)+5(x-25)$

$= (x-25) (x + 5)$

$= (p+q-25) p+q + 5)$

## Factorize : (a2-2a)2-23 (a2-2a)+120.

$a^{2}-2a$

$x^{2}-23x+120 = (x-15) (x-8)$

=$= (a^{2}-2a-15) (a^{2}-2a-8)$

$= a^{2}-2a-15 =a^{2}-5a+3a-15$

$= (a-5)(a+3)$

$= a^{2}-2a-8=(8-4)(a+2)$

$= (a^{2}-2a)^{2}-23 (a^{2}-2a)+120$

$= (a-5) (a+3) (a-4) (a+2)$

## Factorize : (x2-3x)2-8(x2-3x)-20.

$x^{2}-3x=y$

$y^{2}-8y-20=y^{2}-10y+2y-2$

$= y(y-10)+2(y-10)$

$= (y-10)(y+2)$

$= (x^{2}-3x-10)(x^{2}-3x+2)$

$= (x^{2}-5x+2x-10)(x^{2}-2x-x+2)$

$=[x(x-5)x+2(x-5)][(x-1)(x-2)]$

$=(x-1)(x-2)(x+2)(x-5)$

## Simplify and factorize (a+b+c)2-(a-b-c)2+4b2-4c2

$[(a+b+c)^{2}-(a-b-c)^{2}]+[4b^{2}-4c^{2}]$

$=(a+b+c+a-b-c)(a+b+c-a+b+c)+(2b)^{2}-(2c)^{2}$

$=2a\times (2b+2c)+(2b-2c)(2b+2c)$

$= (2b+2c)+(2a+2b-2c)$

$= 2(b+ c) \times2 (a+ b- c)$

$= 4(b+ c) (a+ b- c)$

## Factorize (m+2n)2+101(m+2n)+100

$(m+2n)^{2}+101(m+2n)+100$

$=(m+2n)^{2}+(1+101)(m+2n)+100$

$=(m+2n)^{2}+(m+2n)+100(m+2n)+100$

$=(m+2n) [(m+2n)+1] 100[(m+2n)+1 ]$

?$=(m+2n+1) [ m+2n+100]$

$=(m+2n+1) [ m+2n+100]$

## Factorize : 125a3-27b3+75a2b-45ab2

$125a ^{3}-27b ^{3}+75a ^{2}b-45ab^{2}$

$=125a ^{3}+27b ^{3}-75a ^{2}b-45ab^{2}$

$= 25a ^{2}(5a+3b)-9b^{2}(5a+3b)$

$= (5a+3b) (25a^{2}-9b^{2})$

$= (5a+3b) ( 5a^{2}-3b^{2})$

$= (5a+3b) ( 5a+3b)(5a-3b)$

## Factorize : x4+2x3y-2xy3-y4.

$x^{4}+2x^{3}y-2xy^{3}-y^{4}=(x^{2})^{2}-(y^{2})^{2}+2x^{3}y-2xy^{3}$

$= (x^{2}-y^{2}) (x^{2}+y^{2})+2xy(x^{2}-y^{2})$

$= (x^{2}-y^{2}) (x^{2}+y^{2}+2xy )$

$=(x-y)(x+y)(x+y)^{2}$

$=(x-y)(x+y) ^{3}$

## Factorize $\left ( \frac{1}{64}x \right )^{3} -8y^{3} +\frac{3}{16}x^{2} y-\frac{3}{2}xy^{2}$

$p(x)=\left ( \frac{1}{4}x \right )^{3}-(2y)^{3}+\frac{3}{4}xy \left [ \frac{1}{4} x- 2y\right ]$

$\left ( \frac{1}{4}x -2y\right ) \left [ \left ( \frac{1}{4x} \right ) ^{2}+(2y)^{2}+\frac{1}{4}x\times 2y\right ]$

$+\frac{3}{4}xy [\frac{1}{4}x-2y]$

$=\left ( \frac{1}{4}x -2y\right ) \left ( \frac{1}{16}x^{2}+4y^{2}+\frac{1}{2}xy+\frac{3}{4}xy \right )$

$=\left ( \frac{1}{4}x -2y\right )\left ( \frac{x^{2}}{16}+4y^{2}+\frac{5}{4}xy \right )$

$\left ( \frac{x}{4}-2y \right )\left ( \frac{x^{2}}{16}+\frac{1}{4} xy+xy+4y^{2}\right )$

$=\left ( \frac{x}{4}-2y \right )\left [ \frac{x}{4} \left ( \frac{x}{4}+y \right ) +4y\left ( \frac{x}{4}+y \right )\right ]$

$=\left ( \frac{x}{4}-2y \right ) \left ( \frac{x}{4}+y \right )\left ( \frac{x}{4}+4y \right )$

## Ram has two rectangles having their areas as given below: (1) 25a2-35a+12                   (II) 35y2+13y-12 (i) Given possible expressions for the length and breadth of each rectangle. (ii) Which mathematical concept is used in this problem? (iii) Which value is depicted in this problem

(i) Possible length and breadth of the rectangle are the factors of its given area.

Area    =  $25a^{2}- 35a + 12$

$= 25a^{2}-15a - 20 a + 12$

$=5a (5a-3) -4 (5a-3)$

$= (5a-4) -4 (5a-3)$

So possible length and breadth are (5a -3) and (5a - 4) units, respectively.

(ii)  Area  = $35y^{2}+13y-12$

$= 35y^{2}+28y-15y-12$

$= 7y (5y+4)-3(5y+4)$

$= ( 7y - 3) (5y+4)$

So, possible length and breadth are (7y-3) and (5y+4) units respectively.

(ii) Factorization of polynomials

(iii) Expression of one's desires and news is very necessary.

## The general form of linear equation in two variables is......................

$ax+by+c = 0$ where a,b, c are real numbers and both  $a,b\neq 0.$

## Is  $0x+0y+c = 0$ a linear equation

No, $(\because a,b\neq 0)$ for a linear equation

True .

## In a one day cricket match Raina and Dhoni scored 198 runs. Express this as a linear equation in two variables .

Let the scored by Raina and Dhoni are x & y respectively then

x+y  = 198

## Total number of legs in a heard of goats and hens is 40. Represent this in the form of linear equation of two variables.

Let the number of goats and hens in heard are x & y respectevely them

4x+2y = 40

x- 6y = 5

## Does the following equation x = 5y represent a straight line passing through a point  (0,0)

Given equation  x= 5y

The point  (0, 0 )  statifies the given equation, hence answer is yes .

## Is x=4, y = 0, the solution of y - 4 = 0 ?

No $(\because 0-4\neq 0)$

## The equation x = 7 in two variables can be written as  ...............................

$1.x+0.y=7$

## If the linear equation has solutions (-5, 5) , (0,0) (5-5) then the equation is ..............

$x+y = 0$

## Find the point where equation intersects  3 x +2 y=12  y - axis

As the line intersects y- axis, put x = 0 in the given equation, we get

$3(0)+2y=12$

$\therefore y = 6$

The required point is (0,6).

## If $\sqrt{3x} = \sqrt{2}x+1$ then x is equal to :..................

Given, $\sqrt{3x} = \sqrt{2}x+1$

$\therefore x(\sqrt{3 } - \sqrt{2} ) = 1$

$\therefore x\frac{1}{\sqrt{3 } - \sqrt{2} }$

## If the point (2,3) lies on the line 4y = ax + 5, then a  = ...........................

Given point lies on the

ie, $4(3)= a (2)+5$

$\Rightarrow 2a = 12-5$

$\Rightarrow a = \frac{7}{2}$

## The value of y at x=-1 in the equation 5y =2, is ...................

$\frac{2}{5}$.

## If (0,2) is a solution of the linear equation 2x+3y = k, then find the value f k.

$\because (0,2)$ is the solution of given equation

$\therefore$it satisfies the equation

$\therefore 2(0)+3(2)= k$

$k =6$

## The equation of a line on which the point (6,2) lies is..........

$2x-3y = 6$

(1,-1)

## x = 3, y = 2 is a solution of the linear equation ............

$2x+3y = 12$

(0,0)

x-y = 3

## If  $x = 1$ , $y = -1$ is a solution of the equation  $px-2y=10,$   the value of p is..................

$p=8$

## Any solution of a linear equation $2x+0y+9=0$  in two variables is .........

$\left ( -\frac{9}{2}, m \right )$

## Find three solutions of the linear equation $7x-5y = 35$  in two variables

So, when $y = \frac{7x-35}{5}$

 x 5 0 10 y 0 -7 7

## Give equations of two lines on the same plane which are intersecting at point (2,3).

The equation of two lines on the same plane which are intersecting at point (2,3) are :

$x+ y = 5$

$y - x = 1$

## After 5 years the age of father will be two times the age of his son. Write a linear equation in two variables to represent this statement.

Let father's present age = x years

Son's present age = y years

After 5 years father's age will be = (x+5) years

After 5 years son's age will be = (y+5) years

According to the question, $x+5 =2(y+5)$

$\Rightarrow x+5 =2y+10$

$\Rightarrow x-2y =10-5$

$\Rightarrow x-2y =5$

## A part of the monthly expense of a family on milk is fixed which is Rs. 700 and the remaining varies with the quantity of milk taken extra at the rate of Rs. 25 per litre. Taking the quantity of milk required extra as x litre and total expenditure on milk is Rs. y. Write a  linear equation representing the above information.

According to the question,

$700+25x = y$

$25x-y + 700= 0$

## Check  which of the following is (are)  the solution (s) of the equation  3y - 2x = 1.   (i) (4,3)             (ii) $2(\sqrt{2}, 3\sqrt{2})$

Put x = 4 and  y = 3

then $3y-2x=3(3)-2(4)=1$

So, (4,3)  is the solution of the equation

Again put $x = 2\sqrt{2}$

and $y = 3\sqrt{2}$ ,  then

$3y-2x =3(3\sqrt{2})-2(2\sqrt{2})$

$=5\sqrt{2}\neq 1$

So, $(2\sqrt{2},3\sqrt{2})$ is not a solution of the given equation

## Find the value  of k for which the point (-1,3) lies on the graph  of the equation 2x-y+k = 0

(-1,3) lies on the graph $2x-y+k =0$

$\therefore 2(-1)-3+k=0$

$\Rightarrow k =2+3$

$\Rightarrow k =5$

## If x = 2 and  y = 1 is the solution of the linear equation $2x+3y+k = 0$ find the value of k.

$2x+3y+k =0$

If x = 2 and y = 1 is the solution of the linear equation $2x+3y+k =0$ then

$2(2)+3(1)+k =0$

$\rightarrow k =-7.$

## Express y in terms x from the equation 3x+2y = 8 and check wether the point (4,-2) lies on the line.

$2y=8-3x$

$y =\frac{ 8-3x}{2}$

For $x = 4$

$y = \frac{8-3\times 4}{2}$

$= \frac{8-12}{2}=\frac{-4}{2}=-2$

$\therefore (4,-2)$ lies on the line

## Express y in terms of x in equation 2x-3y =12. Find the points where the line represented by this equation cuts x-axis and y-axis.

Equation $2x-3y=12$

$3y = 2x-12$

$\therefore y = \frac{2x-12}{3}$

On x-axis y=0

$\therefore \frac{2x-12}{3}=0$

$\Rightarrow x = 6$

At point (6,0) the given line cuts the x-axis. On y-axis

x = 0

$\therefore y =\frac{2\times 0-12}{3}$

$\Rightarrow y = -4$

## The point (3,4) lies on the graph of the equation 3y = ax+7. Find the value of 'a'

If points (3,4) lies on

$3y = ax+7$

$\therefore 3 \times 4 = 3a+7$

$\Rightarrow 3a = 12-7 = 5\Rightarrow a\frac{5}{3}$

## Find the co-ordinates of points where the graph of the equation $4x+3 y = 12$ intersects x=axis and y-axis.

Equation : 4x+3y = 12

For intersection with x-axis

y =0

$\therefore 4x=12$

or $x =3$

$\therefore$ co-ordinates are (3, 0)

For intersection with y-axis

$x = 0$$\therefore 3y = 12$

$y=4$

$\therefore$ Co-ordinates are (0,4).

## Find the value of k so that x= -1 and y = -1 is a solution of the linear equation $9kx+12ky =63$

Substituting $x = -1$ and y = -1 in $9kx+12ky =63.$

We get,

$\Rightarrow 9k (-1)+12k(-1)=63$

$\Rightarrow -9k-12k=63$

$\Rightarrow -21k =63\Rightarrow k =-3$

## Find the point at which the equation $3x-2y=6$ meets the x -axis

On the x-axis, y coordinate is zero.

So, put y = 0 in

$3x-2y=6$

we get,

$3x-0=6$

$\rightarrow x =\frac{6}{3}=2$

$\therefore 3x-2y=6$ meets the x-axis at  (2,0)

## If the point (2k-3, k+2)  lies on the graph of the equation 2x+3y+15 = 0, find value of k.

Putting $x =2k-3, y =k+2 in 2x+3y+15 = 0$ We get

$2(2k-3)+3(k+2)+15=0$

$\Rightarrow 4k-6+3k+6+15 =0$

$\Rightarrow k =\frac{-15}{7}$

## Express y in terms of x, given that 2x - 5y =7. Check wether the point (-3,-2) is on the given line.

Given,

2x-5y = 7

$\Rightarrow 5y=2x-7$

$\Rightarrow y =\frac{2x-7}{5}$

Now, when  $x =-3,y=\frac{2(-3)-7}{5}=-\frac{13}{5}\neq -2$

Hence, the point (-3,-2) does not lie on the given line.

## Give the equation of a line passing through (2,14). How many more such lines are there. Write the equation in the form $ax+by+c = 0.$

The line passing through (2,14) is

2y = 14x

y = 7x

In finitely many lines are there.

The equation in the form ax+by+c = 0 is 7x-y+0 = 0.

## Write the equation $y\sqrt{3}=8x+\sqrt{3}$ in the form  $ax+by+c = 0.$check weather  (0,-1) and $(\sqrt{3},9)$ are the solution of this equation

$y =\sqrt{3}=8x+\sqrt{3}$

$\Rightarrow 8x-y \sqrt{3}+\sqrt{3}=0$

Putting x = 0,  y = -1

$\Rightarrow \sqrt{3}+\sqrt{3}\neq 0.$

$\therefore (0,-1)$  is not the solution of the given equation

Putting $x =\sqrt{3} , y=9$

$\Rightarrow 8\sqrt{3} -9\sqrt{3}+\sqrt{3}=0$

Which is correct

$\therefore (\sqrt{3},9)$ is a solution   of the given equation

## When5 times the larger of the two numbers is divided by the smaller, the quotient and the remainder is 2 and 9 respectively. Form a linear equation in two variables. Write it in standard form

Let larger number be x, then 5 times of larger number = 5x and smaller number be y

Qutiont = 2 and remainder = 9

So, according to the question,

$5x= 2y+9$

$\Rightarrow 5x- 2y-9 = 0.$

## ABCD is a square. Co-ordinates of A and C are (-1,-1) and (1,1) respectively. Write the coordinates of B and D. Also write the equations of all the sided of the square.

Given,  A (-1,-1) and C (1,1)

Then,   B (1,-1) and D (-1,1)

Also, equation of sides of square are

AB :  y = -1

BC :  X = 1

CD:  Y=1

DA:  X = -1

## If the point (-1,-5) lies on the graph of 3x = ay+7, then find the value of 'a'

$\therefore$(-1,-5) lies on the graph of

$3x = ay+7$

$\therefore 3(-1)= a\times (-5)+7$

$\Rightarrow 3(-1)=a\times (-5)+7$

$\Rightarrow -3=-5a+7\Rightarrow 5a = 10$

$\Rightarrow a = 2.$

## Given the equation 2x+y= 7, (i)  What is the value of x, when the value of y is 3 (ii) What is the value of y, when the value of x is 4? (iii)Find the one more solution for the above equation ?

(i) when y =3 then

$2x+3 = 7$

$\Rightarrow 2x = 4$

$\Rightarrow x = 2$

(ii) When x = 4, then

$2(4)+y = 7$

$\Rightarrow y=7-8=-1$

(iii) When x = 1, then

$2+y = 7 \Rightarrow y = 5$

$\therefore$ One more solution is (1,5)

## For what value of k, the linear equation 2x+ky =8 has x = 2 and y = 1  as its solution  if x=4, then find the value of y.

The linear equation is 2x+ky = 8

At  x = 2, y = 1

$2(2)+k(1)=8$

$\Rightarrow 4+k=8$

$\therefore k =4$

If x = 4 then

$\Rightarrow 2(4)+4y=8$

$\Rightarrow 8+4y=8$

$\Rightarrow 4y =0$

$\therefore y = 0$

## Show that the points  A (1,2) , B (-1,-16) and C (0,-7) lie on the linear equation  y = 9x-7.

The equation is  y =9x-7

A (1,2)             2 = 9 (1)-7

2 = 2 True

B (-1,-16)       -16= 9(-1)-7=-9-7

= -16 : True

C (0,-7)          -7 = 9(0)-7

= 0-7 = -7 True

## Find the coordinates of the points where the line representing the equation $\frac{x}{4} = 1-\frac{y}{6}$ cut the x-axis and the y-axis.

Writing in the standared form

$3x+2y-12 = 0$

On x-axis, y = 0 $\Rightarrow$

$3x- 12 = 0$

$x = 0$

$\therefore$ Point on the x-axis = (4,0)

On y-axis, x = 0 $\Rightarrow$

$2y-12 = 0$

$y = 6$

$\therefore$ Point on the y-axis = (0,6).

## For what value of p; x =2,y=3 is a solution of (p+1)x-(2p+3) y - 1 = 0 and write the equation

Given equation is

$(p+1)x-(2p+3)y-1=0....(i)$

If x = 2, y = 3  is  the solution of the equation (i) then

$(p+1)2-(2p+3)3-1=0$

$\Rightarrow 2p+2-6p-9-1=0$

$\Rightarrow -4p-8=0$

$\Rightarrow p =-2$

Put the equation (i) then

$-x+y-1=0$

$x-y+1 =0$  is the required equation

## Determine the point on the graph of the linear equation x+ y = 6, whose ordinates is 2 times its abscissa.

Given, y = 2x

Putting y = 2x in the  equation x+y = 6, we get

$x+2x= 6\Rightarrow 3x = 6\Rightarrow x = 2$

$\therefore y = 2\times 2 = 4$

$\therefore$ Required points is (2,4).

## If the point (3,4) lie on the graph of the linear equation 3y = kx+7, then find the value  of k. Also find two more solutions of the equation

On putting (3,4) in the equation of the line

3y = kx +7

$\Rightarrow 3(4) = k(3)+7$

$\Rightarrow 12= 3k+7$

$\Rightarrow 3k = 5$

$\therefore k = \frac{5}{3}$

The equation becomes

9y=5x+21

Two more solutions are (-6,-1) and (12,9)

## Find k in each case if x = 2, y = 1 is a solution o f the equation :  (i) 3x+2y = k,            (ii) 2x-ky= 6 (iii)$\frac{x}{4}+\frac{y}{3} = 5k$

Given,

3x+2y = k

Put x = 2, y = 1 then

$3(2)+2(1)= k \rightarrow k = 8$

(ii) $2x-ky = 6$

Put x = 2, y = 1 then

2(2)-k (1) = 6

$\Rightarrow 4-k= 6\Rightarrow k = 4-6=-2$

(iii) $\frac{x}{4}+\frac{y}{3} = 5k$

Put x = 2, y = 1  then

$\frac{2}{4}+\frac{1}{3} = 5k$

$5k =\frac{10}{12} = \frac{5}{6}$

$\Rightarrow k =\frac{1}{6}$

## Find three different solutions for the equation 3x+2y =1.

Given  $3x+2y = 1$

$\Rightarrow \frac{1-3x}{2}= y$

(i) Put x = 1, then   $y =\frac{1-3(1)}{2}=\frac{-2}{2}=-1$

(ii) Put x = 3 then  $y =\frac{1-3(3)}{2}=\frac{1-9}{2}=-4$

(iii) Put x = 5, then $y =\frac{1-3(5)}{2}=\frac{1-15}{2}=-7$

Hence three different solutions for the equations 3x+2y = 1

 x 1 3 5 y -1 -4 -7

y = k

y = k.

y- axis

infinite

(0,4)

a>0

Parallel, 6

## The graph of the linear equation 3x+5y =15 cuts the x-axis at the point.................

x = 5, y = 0. ie., (5,0)

y = 0

3x-4y+2 = 0

## Express  x in terms of y, in the equation 7x-3y = 15. Check if the line represented by the equation intersects the y-axis at y = -5.

Given

$7x-3 y = 15$

$\Rightarrow x = \frac{15+3y}{7}$

At y- axis  x= 0

$\therefore 7(0)-3y=15$

$\Rightarrow 0-3y = 15$

$\therefore y = \frac{15}{-3} = -5$

Given line intersects the y-axis at y = -5

## Find two different solutions of the equation 4x+3y = 12 from its graph .

Given equation

4x+3y = 12

Put x= 0, then 0+3y = 12

$\Rightarrow y =4$

Hence point on y-axis is (0,4)

Now put y = 0, then

4x+0 = 12

x = 3

Hence the point on x-axis is (3, 0)

## The cost of a toy elephant is the same as the cost of 3 balls . Express the statement as a linear equation in two variables and plot the equation on a graph paper.

Let the cost of a toy elephant =x, ball = y

$\therefore 3 y = x\Rightarrow y = \frac{x}{3}$

 x 3 6 9 y 1 2 3

## Draw the graph of 2x-3y-12 = 0 on the  graph paper

$2x-3y-12 = 0\Rightarrow y =\frac{2x-12}{3}$

 x 6 9 3 y 0 2 2

## Solve the equation $\frac{x}{3}+2=2x-3$ and represent the solution on the   cartesian plane.

$\frac{x}{3} +2 = 2x-3$

$\Rightarrow \frac{x}{3} -2x=-3-2$

$\Rightarrow x -6x=-5 \times 3$

$\Rightarrow -5x=-15$

$\Rightarrow x = 3$

Put x = 3 on the Cartesian plane

On Cartesian plane, x = 3 is a parallel to y-axis

## Write three solutions of the equation 3x = y +3 Draw its graph and find the points where the graph intersects the axes.

$3x=y+3.$three solutions are $x = 1, y = 0: x = 2, y = 3 and x = 0 , y = -3$

From graph it is clear that line meets x- axis  at (1,0) and y- axis at (0,-3)

## If zeroes  of the polynomial $x^{2}+4x+2a$  are $\alpha$ and $\frac{2}{\alpha }$ , then find the value of a.

Products of roots (zeroes)

$= \frac{2a}{1}=\alpha .\frac{2}{\alpha }$

$\Rightarrow\: \: \: \: \: \: \: \: 2a = 2$

$\Rightarrow\: \: \: \: \: \: \: a =1$

## Write the equation  $\frac{x}{2}+\frac{3y}{5}$ in  standared form  and draw the graph

The given equation can be written as,

$5x+6y = -10$

.$\rightarrow - y =(5x+10)/6$

 x -2 4 10 y 0 -5 -10

## ABCD  is a rectangle. Write the equation of its sides. Also find its area.

Equation of the sides are,

AB :   Y = 0

BC:    X = -1

CD:   Y = -4

DA:   X = -4

Area  = 4 x 3

= 12 sq.units

## Draw a triangle whose sides are represented by x = 0, y = 0 and x+y = 3in cartesian system. Also find the co-ordinates of its verticles.

Given,

x+y = 3

On the y axis put x= 0 then

Hence on the y- axis co-ordinates of B is (0,3)

On the x-axis put y = 0then

x = 3

Hence on the x-a xis co-ordinate of Ais (3,0)

Hence a triangle whose sides are x = 0, y = 0, and x+y = 3 is as shown in fig.

Again the vertices of the triangle are A (3,0), B (0,3) and O (0,0).

## Two friends sita and gita, together contributed Rs . 200 towards prime minister relif fund. Write a linerar equation which satisfies this data. Draw the graph.

Let sita contribute = x and contribute= y

According to the question,

x+ y = 200

y = 200-x

 x 0 200 100 y 200 0 100

## Give the geometric representation of  y = 4  as an equation in : (i) one variable  (ii) Two variable

(i) one variable

(ii) Two variable

$0x+y = 4$

$\Rightarrow y -4 = 0$

$\Rightarrow y=0$

The points are (0,4) and (1,4)

## The cost of a pen Rs. 16. Taking a number of pens bought as x and total cost as y, from a linear equation in x and y and draw its graph. Find the cost of 6 pens from the graph

Total cost of value of pen  x number of pens

$\therefore$y = 16x

Table of value of (i) is

 x 0 1 2 3 4 5 6 y 0 16 32 48 64 80 96

From the graph, cost of 6 pens = Rs . 96.

## Give the geometrical representation of the equation 3x+15 = 0  as an equation  (i) One variable  (ii) In two variable

3x + 15 = 0

X = -5

(i) Equay=tion in one variable (Number line ): A point P at a distance of 5 units to left of O on the number line.

(ii) In two variables (Cartesian plane): A line AB parallelto y-axis at a distance of 5 units to the left of y-axis.

## Draw the graph of linear equation 2x+ y = 8 on cartesian plane. Write the co-ordinates of the points where this line intersects x-axis and y-axis

2x+y = 8

y = -2x+8

 x -1 0 1 y 10 8 6

From the graph it is clear that

Line intersects x- axis at (4,0)

and y-axis at (0,8)

## Solve 5x-2 = 3x-8 and represent the solution  (i) On a  number line  (ii) In the cartesian plane

$5x-2 = 3x-8$

$\Rightarrow 2x = -6$

$\Rightarrow x = -3$

(i) Point P (-3,0)  represent the solution x = -3 on the number line

(ii)Line AB represent the solution in the cartesian plane.

## Draw the graph of the linear equation x+y =7. Verify from the graph that (8,-1) is a solution of the equation x+ y = 7

x+ 7  = 7

y = 7-x

 X 5 7 4 Y 2 0 3

From graph it is clear that (8,-1) lies on the line AB.

Hence (8,-1) is a solution of the given equation.

## Draw the paragraph of  $\frac{2}{3}x-y=2$   and find the points where it  cuts the co-ordinate axes.

$\frac{2}{3}x-y = 2$

$\rightarrow 2x-3 y = 6$

$\rightarrow 2x= 3 y + 6$

$\rightarrow x = \frac{3 y +6}{2}$

(i) When the line cut x- axis then put y = 0

i.e,  2x = 6

x = 3

Hence points is (3,0).

(ii)  When the line cut y-axis then put x = 0

i,e., 3y+6 = 0

y = -2

Hence the point  (0,-2)

 x 0 3 6 Y -2 0 2

## Draw the graph of the linear equation 3x+4 y = 6.Find the points where the line representing the equation 3x+ 4y = 6 cuts the axes of x  and y.

Given equation is 3x+ 4 y = 6

(i) When it cut x- axis then

put y = 0 ie., 3x = 6

x = 2

Hence point on x- axis is (2,0)

(ii) When it cut y-axis then

put x = 0 i.,e , 4y = 6

y = 3/2

Hence the point on y-axis is  $\left ( 0, \frac{3}{2} \right )$

3x + 4 y = 6

$y = \frac{6-3x}{4}$

 X 2 -2 6 Y 0 3 -3

## Fahrenheit (F) and Celcius (C) are two different units of temperature and the relation between them is given by $C = \frac{5}{7} (F- 32)$ Draw the graph for this relation. At what temperature both  units read the same.

$x = \frac{5}{9}(y- 32)$

Let x = y = a

$\therefore a = \frac{5}{9}(a-32)$

$\rightarrow 9a =5a-160$

$\rightarrow a = -40$

$\therefore -40^{0}C =-40^{0}F$

## The auto rikshaw fare in a city is charged RS . 10 for first kilometer and @Rs.4 per kilometer for subsequent distance covered. Write the linear equation to express the above statment. Draw the graph of the linear  equation.

Total distance covered =  x km.

Total fare = y km.

Fare for the first kilometer

Subsequent distance  = (x-1) km

Fare for the subsequent distance = Rs. 4(x-1)

According to the question,

y = 10+4 (x-1)

$\Rightarrow$y = 10+4x -4

$\Rightarrow y =4x +6$

Table of solutions

 X 0 1 -1 Y 6 10 2

## Draw the graph of linear equations x+ y = 10 and 2x-y = 5 and  find the point of intersection

x+y = 10 $\rightarrow$  y = 10-x

 X 0 2 3 4 5 Y 10 8 7 6 5

2x-y = 5      y = 2x -5

 X 0 2 5 Y -5 -1 5

Plot these points on the graph paper.