Question

12) If and are the zeros of the quadratic polynomial f(x) = x2 – 2x – 8, then form a quadratic polynomial whose zeros are 3 and 3 . ​

Answer 1

Correct Question:

If α and β are the zeroes of quadratic polynomial x 2 - 2x - 8 then form a quadratic polynomial whose zeroes are 3α and 3β

Your Answer:

If α and β are the zeroes of quadratic polynomial x 2 - 2x - 8 then

α+β= -b/a = -(-2/1) = 2

and

αβ= c/a = (-8/1) = -8

Now we have to form a Quadratic Equation having zeroes 3α and 3β

So, first sum of zeroes

3α + 3β

= 3(α+β)

= 3(2)

= 6

and product of zeroes

3α.3β

=9αβ

= 9(-8)

= -72

So, the format of Quadratic Equation

kx²-(α+β)x+αβ=0

replacing values

=kx²-(6)x+(-72)      ( k=1)

=x² - 6x -72

Answer 2

Given quadratic polynomial :

• x² - 2x - 8.

We will initially find the roots of the polynomial using middle term splitting method.

=> $\sf{x^2 - 2x - 8 =0}$

=> $\sf{x^2 - 4x + 2x - 8 = 0}$

=> $\sf{x(x-4)+2(x-4) = 0}$

=> $\sf{(x-4) (x+2) = 0}$

=> $\sf{x-4 = 0\;\;\;or\;\;\; x + 2 = 0}$

=> $\sf{x=4\;\;\;or\;\;\;x=-2}$

Now from here we have the two roots of the polynomial.

Let α equals as 4.

Let β equals as -2.

Find the sum and product of α and β turn by turn.

=> $\sf{Sum\: of\: roots\: = 3\:\alpha + 3\:\beta}$

=> $\sf{Sum\: of\: roots \:= 3(4) + 3(-2)}$

=> $\sf{Sum\: of\: roots\: =12 + (-6)}$

=> $\sf{Sum\: of \:roots \:= 6.....(1)}$

Now find the product of the roots.

=> $\sf{Product\: of \:roots \:= 3\:\alpha\times\: 3\:\beta}$

$\sf{=>\:Product\: of \:roots \:= 3(4) \times\:3(-2)}$

$\sf{=>\:Product\: of \:roots \:= 12 \:\times\:-6}$

=> $\sf{Product\: of \:roots \:= -72.....(2)}$

It is known that the quadratic polynomial is of the basic form as :

• $\sf{x^2 - (Sum\:of\:roots)x + (Product\: of\:roots)}$

Plug in the values from equation (1) and (2),

=> $\sf{x^2-6x+(-72)}$

$\boxed{\red{\sf{x^2-6x-72....\big[Required\: Polynomial\big]}}}$