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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.
Answer:-
Given:
α + β = 24 -- equation (1)
α - β = 8 -- equation (2).
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
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