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Relations between Roots and Coefficients : Higher Order Equations

If α,β are ze...

Question

If $α, β$ are zeroes of a polynomial $6 x^{2} + x - 2$ , then the polynomial whose zeroes are $2 α + 3 β$ and $3 α + 2 β$ , is .

A

6 x^{2} + 5 x - 1

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B

6 x^{2} - 5 x + 1

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C

6 x^{2} - 5 x - 1

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D

6 x^{2} + 5 x + 1

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Solution

The correct option is A $6 x^{2} + 5 x - 1$
$y = 6 x^{2} + x - 2 \Rightarrow α + β = \frac{- 1}{6} and αβ = \frac{- 2}{6} = \frac{- 1}{3} Now, Let k_{1} = 2 α + 3 β and k_{2} = 3 α + 2 β ∴ k_{1} + k_{2} = (2 α + 3 β + 3 α + 2 β) = 5 (α + β) = \frac{- 5}{6} and k_{1} \times k_{2} = (2 α + 3 β) \times (3 α + 2 β) = 6 (α^{2} + β^{2}) + 13 αβ = 6 [(α + β)^{2} - 2 αβ] + 13 αβ = 6 (α + β)^{2} + αβ = 6 \times \frac{1}{36} + (\frac{- 1}{3}) = \frac{1}{6} - \frac{1}{3} = - \frac{1}{6} ∴ Required polynomial will be : k [x^{2} - (k_{1} + k_{2}) x + k_{1} k_{2}], where k is non zero real number \Rightarrow k [x^{2} + \frac{5}{6} x - \frac{1}{6}] \Rightarrow \frac{k}{6} [6 x^{2} + 5 x - 1] by putting k = 6, we will get the required answer$

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