Zero is a rational or an irrational number?

about 2 years ago 1 Answer 410 views

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

about 8 months ago 1 Answer 111 views

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

about 7 months ago 2 Answer 101 views

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}+32times 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)

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  

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

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=2alpha +2beta = 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 = 2alpha +2beta = 4alpha beta

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

                          p = -5  and q = -6

 

 

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}.

 

 

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}+2alpha 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}+2alpha 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}

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+2times 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=10times 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

  3times 5+frac{3(y-5)}{3}=3times 0

  15+y-5=0,y=5-15=-10

  Coordinates of Q =(30,-10)