If α and β are the zeroes of a quadratic Polynomial such that α + β = 24 , α - β = 8 . Find a quadratic Polynomial having α and β as it's zeroes.

α + β = 24 -- equation (1)

α - β = 8 -- equation (2).

A quadratic Polynomial having α and β as it's zeroes.

We know that,

(a + b)² - (a - b)² = 4ab

⟹ (α + β)² - (α - β)² = 4ab

Substitute the values from equations (1) & (2).

⟹ (24)² - (8)² = 4αβ

⟹ 576 - 64 = 4αβ

⟹ 512/4 = αβ

⟹ αβ = 128

Now,

General form of a Quadratic equation is x² - (sum of the zeroes)x + Product of the zeroes = 0.

We have:

Sum of the zeroes = 24

Product of the zeroes = 128.

So,

⟹ Required quadratic Polynomial = x² - 24x + 128

(a + b)² - (a - b)² = 4ab

Proof:

⟹ (a + b)(a + b) - (a - b)(a - b) = 4ab

⟹ a(a + b) + b(a + b) - [ a(a - b) - b(a - b) ] = 4ab

⟹ a² + ab + ab + b² - (a² - ab - ab + b²) = 4ab

⟹ a² + 2ab + b² - a² + ab + ab - b² = 4ab

⟹ 4ab = 4ab

I hope it will help you.

Regards.

ANSWER

α+β=24

α−β=8

(α−β)

2

=(α+β)

2

−4αβ=8

2

or,24

2

−4αβ=64

or,4αβ=576−64=512

or,αβ=128

⇒x

2

−(α+β)x+αβ=0

or,x

2

−24x+128=0

Hence x

2

−24x+128=0 is the quadratic polynomial for the required condition

the answer is simply just 6n add the two together as they have an n