Zero is a rational or an irrational number?

find zeros of (x+2) (x-3)

solve for x and y a2/x-b2/y=0,a2b/x+b2a/y=a+b where x and y are not equal to zero

## 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.

Not defined

Binomial

## 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

## 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 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)

$f (0)= b=0$

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

$a=2$

$a=2, b= 0$

## 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

## 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  $p(x)$

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

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

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

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

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

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

## 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, $p (x) = (x-1) (x+3) ( 3x-1)$

## 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,  $[p \left ( -\frac{1}{2} \right ) = 0]$

$p\left ( -\frac{1}{2} \right )=2\times \left ( \frac{-1}{2} \right )^{4}-a \left ( \frac{-1}{a} \right )^{3} +4\left ( \frac{-1}{4} \right )^{3}-\left ( \frac{-1}{2} \right )+2$

$=0$

$\Rightarrow 2\times \frac{1}{16}-a\left ( \frac{-1}{8} \right )+4\left ( \frac{1}{4} \right )-\left ( \frac{-1}{2} \right )+2=0$

$\Rightarrow \frac{1}{8}+\frac{a}{8} +1+\frac{1}{2}+2=0$

$\Rightarrow \frac{29}{8}+\frac{a}{8} = 0$

$\therefore a = -29$

## 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 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

## 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 zeroes  of the polynomial $x^{2}+4x+2a$  are $\alpha$ and $\frac{2}{\alpha }$ , then find the value of a.

Products of roots (zeroes)

$= \frac{2a}{1}=\alpha .\frac{2}{\alpha }$

$\Rightarrow\: \: \: \: \: \: \: \: 2a = 2$

$\Rightarrow\: \: \: \: \: \: \: a =1$

## Check whether 4n can be  end with the digit  0  for any natural number n.

If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5.That is, the prime factorization of 4n would contain the prime  5. This is not possible  because 4n = (2)2n ; so the only prime in the factorization of 4n is2. So, the uniqueness of the Fundamental Theorem of arithmetic guarantees that there are no other primes in the factorization of 4n. So there is no natural number  n for which 4n ends with the digit zero

## Show that 7n cannot end with the digit zero, for any natural number n.

7n =(1x 7)n = 1n x 7n

So the only prime in the factorization of  7n is  7, not 2 or  5.

7n cannot end will the digit zero.

## Check wether (15)n can end with digit 0 for any   $n \euro N.$

(15)n can end with the digit 0 only if (15)n  is divisible by 2 and  5.

But prime factors of (15)n are 3n x 5n

By Fundamental theorem of arithmetic, there is no natural number n for which (15)n ends with the digit zero

## Find the HCF 256 and 36 using Euclid's division Alogaritham. Also find their LCM and verify that HCF and LCM = Product of two numbers.

Since 256 > 36, we state with 256 as dividend and 36 as divisor, we have

256 = 36 x 7 +4

Now, 4 is the remainde, which is not zero, so we again apply Euclid Division Alogaritham to  36 and 4 we have

36 = 4 x 9 +0

Now, remainder has become zero and 4 is the divisor.

Hence the HCF of 256 and 36 is 4

LCM

256 = 28

36 = 22 x 32

LCM  (36,256) = 28 x 32 = 256 x 9

= 2304

HCF x LCM  = Product of the two number

4 x 2, 304 = 256 x 36

9216 = 9,216

## Find the quadratic polynomial whose sum and product of the zeroes are $\frac{21}{8}$ and  $\frac{5}{16}$ respectively.

According to the question,

Sum of zeroes  = $\frac{21}{8}$

and product of zeroes  = $\frac{5}{16}$

= $x^{2}-$(Sum of zeroes )$x$ + product of zeroes

$= x^{2}-\left ( \frac{21}{8} \right ) x +\frac{5}{16}$

$=\frac{1}{16}(16x^{2}-42x +5)$

$=(16x^{2}-42x +5)\frac{1}{16}.$

## If 'm' and 'n' are the zeroes of the polynomial  $3x^{2}+11 x-4,$ find the value of $\frac{m}{n} + \frac{n}{m}$

Let, $p(x)= 3x^{2}+11x -4$

$=3x^{2}+12x-x-4$

$=3x (x+4)-1 (x+4)$

$= (3x-1) (x+4)$

So, zeroes are, $m = \frac{1}{3}$   and  $n = -4$

Now,   $\frac{m}{n}+\frac{n}{m}= \frac{\left ( \frac{1}{3} \right )}{-4}+\frac{-4}{\left ( \frac{1}{3} \right )}=\frac{1}{-12}-12$

$= \frac{-145}{12}$

## If  $\alpha$  and $\beta$  are the zeroes  of a polynomial   $x^{2}-4\sqrt{3}x+3,$  then find the value of  $\alpha +\beta -\alpha \beta$

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

If  $\alpha$  and  $\beta$  are the zeroes  of  $x^{2}-4\sqrt{3}x+3$

then $\alpha + \beta =-\frac{b}{a}$

$\Rightarrow \alpha + \beta = -\frac{(-4\sqrt{3})}{1}$

$\Rightarrow \alpha + \beta = 4\sqrt{3}$

and        $\Rightarrow \alpha \beta = \frac{c}{a}$

$\Rightarrow \alpha \beta = \frac{3}{1}$

$\Rightarrow \alpha \beta = 3$

$\therefore \alpha + \beta -\alpha \beta = 4\sqrt{3}-3$

## If one of the zeroes of the quadratic polynomial is $f (x) = 14x^{2} -42k^{2}x -9$ is a negative of the other, find the value of 'k'.

Given

Let one zero  be $\alpha$,

$\therefore$ The other  = $-\alpha$

$\therefore$ Sum of zeroes = $\alpha +(-\alpha ) = 0$

Sum of zeroes  = $-\frac{Coefficient of x}{Coefficient of x^{2}}$

According to the equation,

Sum of zeroes  = $\frac{42k^{2}}{14}=3k^{2}$

$3k^{2} = 0 \rightarrow k = 0$

## If the zeroes of the polynomial $x^{2}+px+q$   are double in value to the zeroes of  $2x^{2} -5x -3$ , find the value of  p and q.

Let, $f (x) = 2x^{2}-5x -3$

Let the zeroes of polynomial be $\alpha$  and $\beta$ then,

Sum of zeroes   $=\alpha+ \beta = \frac{5}{2}$

$product\: \: of\: \: zeros=\alpha \beta = -\frac{3}{2}$

According to the question, zeroes of  $x^{2}+px+q$ are $2 \alpha$ and $2 \beta$

Sum of zeroes  = $\frac{coeff. of x}{coeff.of x ^{2}} = \frac{-p}{1}$

$-p=2\alpha +2\beta = 2 (\alpha +\beta )$

$-p=2 \times \frac{5}{2} = 5 \Rightarrow p=-5$

Product of zeroes  =$\frac{constant\: \: term}{coeff.\: of \: x ^{2}} = \frac{q}{1}$

$\Rightarrow q = 2\alpha +2\beta = 4\alpha \beta$

$\Rightarrow q = 4 \left ( -\frac{3}{2} \right ) = -6$

$p = -5$  and q = -6

## If the sum and product of the zeroes of the polynomial  $ax^{2}-5x +c$ is equal to 10 each, find the value  of 'a' and  'c'

Given, Polynomial $f (x) =ax^{2}-5x+c$

Let the zeroes  of  $f (x)$ are $\alpha$ and  $\beta$ , then according to the question

Sum of zeroes,  ($\alpha + \beta$)  = Product of zeroes, $(\alpha \beta )$ = 10

Now,  $\alpha + \beta = - \frac{Coefficient\:\: of \:\: x }{ Coeff \:\: of\:\: x^{2}} =- \frac{-5}{a}$

$\Rightarrow 10 = \frac{+5}{a}$

$\therefore a =\frac{1}{2}$

and      $\alpha \beta = \frac{Constant\: \: term}{Coeff.\: \: of\: \: x ^{2}}$

$\Rightarrow 10 = 2c$

$\therefore c = 5$

Hence  $a = \frac{1}{2}\: \: and\: \: c = 5$



## If  $\alpha$   and  $\beta$ are zeroes of the polynomial  $p(x)=3x^{2}-4x-7$ then form a quadratic polynomial whose zeroes are  $\frac{1}{\alpha }$  and  $\frac{1}{ \beta }$

Given, p (x) = 3x2 -4x-7 and $\alpha$ and $\beta$  are its zeroes.

$sum\: \: of\: zeroes=\alpha +\beta = \frac{coefficient\: \: of\: \: x}{coefficient\: \: of\: \: x^{2}}$

$= -\left ( -\frac{4}{3} \right )= \frac{4}{3}$

$product\: \: of\: \: zeroes= \alpha \beta = \frac{Constant\: \: term}{ Coefficient of x^{2}}$

$= \left ( \frac{-7}{3} \right ) = -\frac{7}{3}$

For the new polynomial

$sum\: \: of\: \: zeroes\: \: =\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{\frac{4}{3}}{-\frac{7}{3}}=\frac{-4}{7}$

$product\: \: of\: \: zeros= \frac{1}{\alpha }\times \frac{1}{ \beta }=\frac{1}{\alpha \beta }=\frac{1}{\frac{-7}{3}}=\frac{-3}{7}$

Required quadratic polynomial = x2 - ( Sum of zeroes ) x + Product of zeroes

$= x^{2}- \left ( \frac{-4}{7} \right ) x +\left ( \frac{-3}{7} \right )$

$= \frac{1}{7} (7x^{2}+4x-3)$

$=( 7x^{2}+4x -3) \frac{1}{7}.$

## Show that  $\frac{1}{2}$  and   $\frac{-3}{2}$ are the zeroes of the polynomial $4x^{2}+4x-3$ and verify the  relationship between  zeroes and  coefficients of the polynomial.

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

$\Rightarrow f\left ( \frac{1}{2} \right ) = 4 \left ( \frac{1}{4} \right )+4 \left ( \frac{1}{2} \right )-3$

=1+2-3 =0

and  $\Rightarrow f\left ( -\frac{3}{2} \right ) = 4 \left ( \frac{9}{4} \right )+4 \left (- \frac{3}{2} \right )-3$

= 9 - 6 -3 = 0

$\therefore \frac{1}{2}, - \frac{3}{2}$  are zeroes of polynomial $4x^{2}+4x-3.$

$sum\: \: of\: \: zeroes\: =\frac{1}{2}-\frac{3}{2}=-1 =\frac{-4}{4}$

$=- \frac{Coefficient \: of \: \: x}{Coefficient \: \: of \: \: x^{2}}$

$product\: \: of\: \: zeroes\: \: =\left ( \frac{1}{2} \right )\left ( -\frac{3}{2} \right )= \frac{-3}{4}$

$= \frac{Constant\: \: term }{Coefficient \: \: of\: \: x^{2}}$

$\therefore$ Relation between zeroes and coeff. of polynomial is verified.

## Polynomial  $x^{4}+7x^{3}+7x^{2}+px +q$  is exactly divisible by  $x^{2}+7x+12$, then find the value of p and q.

Factors  of $x^{2}+7x+12:$

$x^{2}+7x+12 = 0$

$\Rightarrow x^{2}+4x+3x+12 = 0$

$\Rightarrow x (x+4)+3( x+4)=0$

$\Rightarrow (x+4)( x+3)=0$

$\Rightarrow x =-4, -3$       ......................................(1)

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

If p(x) is exactly divisible by $x^{2}+7x+12$, then x = -4  and  x = -3 are zeroes of  p(x) [ from eq (1)]

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

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

But  p (-4) = 0

$\therefore 0 = 256-488 + 112 - 4p +q$

$\Rightarrow 0 = - 4p +q -80$

$\Rightarrow 4p -q =-80$      ....................................(2)

and $p(-3) = (-3) ^{4} + 7(-3) ^{3}+7(-3)^{2}+p (-3)+q$

but p(-3) =0.

$\therefore 0 =81-189+63-3p +q$

$\Rightarrow 0= -3p+q -45$

$\Rightarrow 3p-q = -45$  ........................................(3)

On solving  eq (2) and  eq (3) by elimination method, we get

p =-35

On Putting the value of  p  in eq  (1)

$4(-35)-q= -80$

$\Rightarrow -140-q =-80$

$\Rightarrow -q = 140 -80$

$\Rightarrow -q = 60$

$\therefore q = 60$

Hence, p = -35, q = -60

## If $\alpha$  and  $\beta$  are the zeros of the polynomial  $x^{2}+4x+3,$  find the polynomial whose zeroes  are   $1+\frac{\beta }{ \alpha }$   and  $1+\frac{ \alpha }{ \beta }$

Since $\alpha$ and $\beta$ are the zeroes of the cubic polynomial  $x^{2}+4x+3$

then  $\alpha +\beta = -4$

and  $\alpha \beta = 3$

$sum\: \: of\: \: zeroes\: \: =1+\frac{\beta }{\alpha }+1+\frac{\alpha }{\beta }$

$=\frac{\alpha \beta +\beta ^{2}+\alpha \beta +\alpha ^{2}}{\alpha \beta }$

$=\frac{\alpha^{2}+ \beta ^{2}+2\alpha \beta }{\alpha \beta }$

$=\frac{(\alpha+ \beta )^{2} }{\alpha \beta }=\frac{(-4)^{2}}{3}=\frac{16}{3}$

$product\: \: of\: \: zeroes= \left ( 1+\frac{\beta }{\alpha } \right ) \left ( 1+\frac{\alpha }{\beta } \right )$

$= 1+\frac{\beta }{\alpha }+\frac{\alpha }{\beta }+\frac{\alpha \beta }{\alpha \beta }$

$=\frac{\alpha ^{2}+\beta ^{2}+2\alpha \beta }{\alpha \beta } = \frac{(\alpha +\beta )^{2}}{\alpha \beta }$

$= \frac{(-4)^{2}}{3}=\frac{16}{3}$

But required polynomial = x2 - (Sum of the zeroes)x + product of the zeroes

$= x^{2}-\left ( \frac{16}{3} \right )x+\frac{16}{3}$

$= \left (x^{2} -\frac{16}{3} x+\frac{16}{3} \right )$

$= (3x^{2}-16x+16)\frac{1}{3}$

## If  $\alpha$ and $\beta$ are the  zeroes of the polynomial  $2x^{2}-4x+5$ , find the value of  :   $(1)\: \: \: \alpha ^{2}+\beta ^{2}$  $(2)\: \: \: \frac{1}{\alpha }+\frac{1}{\beta }$  $(3)\: \: \: (\alpha -\beta )^{2}$  $(4)\: \: \: \frac{1}{\alpha ^{2}}+\frac{1}{\beta ^{2}}$ $(5)\: \: \: \alpha ^{3}+\beta ^{3}$

Given   $\alpha\: \: and \: \: \beta$ are the zeroes  of  $2x^{2}-4x+5$

$\rightarrow \alpha +\beta = \frac{-(-4)}{2} = 2$

and  $\alpha \beta = \frac{ 5}{2}$

Now,

$\left ( 1 \right )\: \: (\alpha ^{2}+\beta ^{2}) = (\alpha +\beta )^{2}-2\alpha \beta = 2^{2}-2\times \frac{5}{2}=4-5$

$= -1$

$\left ( 2 \right )\: \: \frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{2}{\frac{5}{2}}=\frac{4}{5}$

$\left ( 3 \right )\: \: (\alpha -\beta )^{2}= (\alpha +\beta )^{2}-4\alpha \beta =2^{2}-\frac{4\times 5}{2}$

$=4-10 =-6$

$\left ( 4 \right )\: \: \: \frac{1}{\alpha ^{2}}+\frac{1}{\beta ^{2}}=\frac{\alpha ^{2}+\beta ^{2}}{(\alpha \beta )^{2}}=\frac{-1}{(\frac{5}{2})^{2}}=\frac{-4}{25}$

$\left ( 5 \right )\: \: \: (\alpha ^{3}+\beta ^{3})=(\alpha +\beta )^{3} -3\alpha \beta (\alpha +\beta )$

$=2^{3}-3\times \frac{5}{2}\times 2=8-15=-7$

## If the squared difference of the zeroes of the quadratic polynomial  $f (x)=x^{2}+px+45$ is equal to 144, find the value of p.

The given quadratic polynomial is  $f (x)= x^{2}+px +45.$ Let $\alpha$ and $\beta$ be the zeroes of the given quadratic polynomial

$\therefore \alpha +\beta = -p$ and $\alpha \beta =45$

Given $(\alpha -\beta )^{2}= 144$

$\rightarrow (\alpha +\beta )^{2}-4\alpha \beta = 144$

$\rightarrow (-p)^{2}-4\times 45 =144$

$\Rightarrow p ^{2}-180 = 144$

$\Rightarrow p ^{2}= 144 +180 = 324$

$\therefore p = \pm \sqrt{324} = \pm 18$

Thus the value of p is $= \pm 18$

## A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digit. Find the number.

Let units digits and tens digits of the two digit number be x and y

$\therefore$  Number  is  10 y + x

According to question,

$10y +x = 4 (y+x)$

$10y +x = 4y+4x$

$10y -4y = 4x-x$

$6y =3x$

$2y =x$

Also,

$10 y + x = 3x y$

$10 y + 2y=3(2y)y$

$12 y =6y^{2}$

$6 y^{2} -12y =0$

$6 y(y-2)=0$

$6y =0$    or y=2

Rejecting y = 0 as the number can not be zero.

$\Rightarrow x = 4$

$\therefore$  Required number is  24.

## In given figure, the graph of a polynomial p(x) is shown. Calculate the numbers of zeroes of p(x).

The numbers of zeroes of p(x) is 1.

## Calculate the zeroes of the polynomial $p(x) = 4x^{2}-12x+9$

Let,

$p(x) = 4x^{2}-12x+9$

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

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

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

The zeroes are  $\frac{3}{2}, \frac{3}{2} \rightarrow x = \frac{3}{2}, \frac{3}{2}$

Hence, zeroes of the polynomial are   $\frac{3}{2}, \frac{3}{2}$

## If sum of the zeroes of the quadratic polynomial  $3x^{2}-kx+6$ is  3, then find the value of k.

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

$Sum \: \: of \: \: the \: \: zeroes = 3 = -\frac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: x^{2}}$

$3 = - \frac{(-k)}{3}$

$k =9$

## If -1 is a zero of the polynomial  $f(x) = x^{2} -7x -8$  then calculate the other zero.

$f(x) = x^{2} -7x -8$

Let the other zero be k, then

$Sum\, \, of\,\, the\, \, zeroes \, -1+k = -\left ( \frac{-7}{1} \right ) = 7$

$k = 8$

## Find all the zeroes of $f(x) = x^{2}-2x$

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

$= x (x-2)$

i,e., $f(x)=0 \rightarrow x = 0 \: \: or \: \: x = 2$

Hence, zeroes are 0 & 2

## Find the zeroes of the quadratic polynomial $\sqrt{3}x^{2}-8x+4\sqrt{3}$

$p (x)= \sqrt{3}x^{2}-8x+4\sqrt{3}$

$= \sqrt{3}x^{2} -6x-2x+4\sqrt{3}$

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

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

$Zeroes\, \, are\, \, x = \frac{2}{\sqrt{3} } , 2\sqrt{3}$

## Find  a quadratic polynomial, the sum  and product of whose zeroes are 6 and 9 respectively  Hence find the zeroes.

Sum of zeroes  = 6, product of zeroes = 9

$\therefore$ Quadratic polynomial is $x^{2}-6x+9$

Also,  $x^{2}-6x+9 = 0$

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

$x = 3, 3$

Hence zeroes are 3,3

## Form a quadratic polynomial p(x) with 3 and 2/5  as sum and product of its zeroes, respectively .

According to the question,

sum of zeroes = 3

Product of zeroes = $- \frac{2}{5}$

$=x^{2}-x$ (Sum of zeroes) +product of zeroes

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

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

$= \frac{1}{5} (5x^{2}-15x-2)$

the quadratic polynomial is $\left ( 5x^{2}-15x-2 \right )\frac{1}{5}$

## If p,q are zeroes of polynomial $f(x)= 2x^{2}-7x+3$ find the value of $p^{2}+q^{2}$

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

$Sum \: \: of \: roots = p+q = - \frac{Coefficient \: of \: x}{Coefficient \: of \: x^{2}}$

$= -\left ( \frac{-7}{2} \right ) = \frac{7}{2}$

$Product \: \: of \: \: roots = pq = \frac{Constant\, \, term }{Coefficient \: of \: x^{2}} = \frac{3}{2}$

We know that,

$(p+q)^{2}=p^{2}+q^{2}+2pq$

$p^{2}+q^{2}= (p +q )^{2}+2pq$

$= \left ( \frac{7}{2} \right )^{2}-3= \frac{49}{4}-\frac{3}{1} = \frac{37}{4}$

## Find the condition that zeroes of polynomial $p(x )= ax^{2}+bx+c$ are reciprocal of each other.

$p(x)= ax^{2}+bx+c$

$Let\, \, \alpha \: \: and \: \: \frac{1}{\alpha }\, \, be\, \, the\, \, zeroes\, \, of\, \, p(x)\, \, then$

$Product\, \, of\, \, zeroes, \alpha \times \: \: \frac{1}{\alpha } = \frac{c}{a}$

So, required condition is, c = a

## Find the values of a and b , if they are the zeroes of polynomial $x^{2}+ax+b.$

$Sum \: \: of\: \: zeroes = -\frac{Coefficient \: \: of \: \: x}{Coefficient \: \: of \: \: x^{2}}$

$\therefore a+ b = -a$

$2a+b = 0$

Product of zeroes = $\frac{Constant \: \: term }{Coefficient \: \: of \: \: x^{2}}$

$\therefore ab = b$

$a = 1$

then  $b = -2$

## A policeman and a thief are equidistant from the jewel box. Upon considering the jewel box as origin, the position of a policeman is (0,5). If the ordinate of the position of thief is zero, then what will be the position of the thief?

The position of thief = (5,0) or (-5,0)

## A point whose y- coordinate is zero and x-coordinate is 5 will lie on  a. y-axis        b.x-axis           c. origin       d. None f theses

(5,0) lies on x-axis

31

## Calculate the coordinates of the point at which the circle of centre (2,4) and radius 4 units cuts the y axis.

Centre of the circle =(2,4). radius =4

Equation of the circle =$(x-2)^2 +(y-4)^2 =4^2$

The coordiantes of the point where the circle cuts the y axis is zero.$(0-2)^2 +(y-4)^2 =16$

$4+(y-4)^2 =16,(y-4)^2 =12$

$y-4=\pm \sqrt{12}=\pm \sqrt{4\times 3}=\pm 2\sqrt{3}$

$y=4\pm 2\sqrt{3}=4+2\sqrt{3},4-2\sqrt{3}$

Coordinates of the point where the circle cuts the y axis =$(0,4+2\sqrt{3}),(0,4-2\sqrt{3})$

## In the figure, equation of the line joining the points A and B is x+2y=10. P and Q are points on this line. a. Find the coordinates of the points A and B. b. Find the coordinates of the points P that divides the line AB in the ratio 2:3. c. If AQ: BQ =2:3, find the coordinates of the point Q.

a. Equation of the line AB is x+2y=10

Since the y coordinates of the point A is zero, $x+2\times 0=10,x=10$

Coordinates of A = (10,0)

Since the x coordinates of the point B is sero, 0+2y=10,y=5

Coordinates of B =(0,5)

b. AP : PB =2:3

x coordinates of P =$x_{1}+\frac{p}{(p+q)}(x_{2}-x_{1})$

$=10+\frac{2}{5}(0-10)=10+\frac{2}{5}\times -10$

$=10+-6$

y coordinates of p =$y_{1}+\frac{p}{p+q}(y_{2}-y_{1})$

$=0+\frac{2}{5}(5-0)=0+2=2$

Coordinates of P=(6,2)

c. AQ: BQ=2:3

Then BA: AQ=1:2

if x is the x coordinates of Q.

$0+\frac{1}{3}(x-0)=10$

$\frac{1}{3}(x-0)=10$

$x-0=10\times 3=3 0,x=30$

If y is the y coordinates of Q.

$5+\frac{1}{3}(y-5)=0,5+\frac{(y-5)}{3}=0$

$3\times 5+\frac{3(y-5)}{3}=3\times 0$

15+y-5=0,y=5-15=-10

Coordinates of Q =(30,-10)

## Check whether x-3 is a factor of the polynomial $2x^3 -x^2 -3x+4$.

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

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

$=54-9-9+4=40$,

since p(3) is not equal to zero, x-3 is not a factor of p(x).