Question

if alpha and beta are the zeros of a polynomial x^2-3x-1 then find a quadratic polynomial whose zeroes are 1/alpha^2 &1/beta^2​

Answer 1

Answer:

x^2 -11x - 1 = 0

Step-by-step explanation:

x^2-3x-1

Here;

 a = 1,   b = -3, c = -1

Finding Sum and Product of the Zeros of equation

α + β = -b/a = -(-3)/1 = 3

    αβ = c/a  = -1/1 = -1

Now if the roots(zeros) are 1/α^2 and 1/β^2

Then Sum 'S' is written as;

  1/α^2 + 1/β^2 = (α^2 +β^2)/(α^2)(β^2) ..........( i )

As we know  (α + β)^2 = α^2 +β^2 + 2αβ

                      (α + β)^2 - 2αβ = α^2 +β^2

Put Value of α^2 +β^2 in equation ( i )

                    =  {(α + β)^2 - 2αβ}/(α^2)(β^2)

                    =  {(α + β)^2 - 2αβ}/(αβ)^2

 Put Values of α + β and αβ as Derived earlier

                    =  {(3)^2 - 2(-1)}/(-1)^2

                    = (9 +2)/1

                    = 11

Now the product 'P' is as follows:

             1/αx1/β = 1/αβ

Put Value αβ here

              1/αβ = 1/-1 = -1

The Equation is written as

             x^2 -Sx + P = 0

Putting values of S and P here

             x^2 -11x + (-1) = 0

             x^2 -11x - 1 = 0 Answer