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If α, β are the zeroes of the polynomial (x2 – x – 12). Then form a quadratic equation whose zeroes are 2α and 2β. |
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Answer» Given that, α and β are zeros of the polynomial x2 – x – 12. The sum of zeros = α + β = \(\frac{-b}{a}\) = \(\frac{−(−1)}{1}\) = 1. The product of zeros = αβ = \(\frac{c}{a}\) = \(\frac{−12} {1}\) = −12. Now, (α – β)2 = (α + β)2 – 4αβ = 12 – 4 × –12 = 1 + 48 = 49. ⇒ α − β = ±7. Case I :- α − β = 7 and α + β = 1. ⇒ 2α = 8 ⇒ α = 4. Then, 4 + β = 1 ⇒ β = 1 − 4 = −3. i.e., α = 4 and β = −3 . Case II :- α − β = −7 and α + β = 1 ⇒ 2α = −6 ⇒ α = −3. Then, −3 + β = 1 ⇒ β = 1 + 3 = 4. i.e., α = −3 and β = 4. Hence, –3 and 4 are zeros of given polynomial x2 – x – 12. We want to find a quadratic equation whose zeros are 2α and 2β i.e., quadratic equation is (x – 2α) (x – 2β) = 0 ⇒ (x − 8)(x + 6) = 0 (∵ either α = −3 & β = 4 or α = 4 and β = −3) ⇒ x2 − 2x − 48 = 0. ∴ Required quadratic equation is x2 − 2x − 48 = 0. |
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