Question

Find a polynomial whose zeroes are squares of the zeroes of the polynomial 3x^2 + 6x - 9

Answer 1

=3x^2 + 6x - 9
=3x^2 + 9x -3x- 9
=3x(x+3) - 3(x +3)
=(x+3)(3x-3)

x = -3 (Ist zero )
x =1 ( 2nd zero )

required zeroes =
= -3^2 = 9
= 1^2 = 2

Sum of zeroes = 9 +2 = 11 = -b /a
Product of zeroes = 9*2 = 18 = c/a
a = 1 ; b =-11 ; c =18

REQUIRED POLYNOMIAL =  x^2 -11x +18

HOPE IT HELPS !!

Answer 2

Sol:
Given polynomial 3x2 + 6x - 9  = 0
⇒ 3x2 + 9x -3x - 9  = 0
3x( x + 3) -3( x + 3) = 0
3x - 3 = 0 , x + 3 = 0
x = 1 , x = - 3.
Given that zeroes are square of the zeroes of the polynomial
so that x = 1, 9
Sum of the roots α + β = 10.
Product of the roots αβ = 9
According to the question x2 -(α + β)x + αβ = 0

 ∴ the polynomial is x2 -10x + 9 = 0.

Hope it helps.