If $α, β$ are the roots of the equation $2 x^{2} - 3 x - 6 = 0$ , then the equation whose roots are $α^{2} + 2$ and $β^{2} + 2$ is

4 x^{2} - 49 x + 118 = 0

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4 x^{2} - 40 x + 100 = 0

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4 x^{2} + 49 x + 108 = 0

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None of these

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Solution

The correct option is A $4 x^{2} - 49 x + 118 = 0$
${2 x}^{2} - 3 x - 6 = 0 α + β = \frac{3}{2} α β = - 3 n e w r o o t s a r e (α^{2} + 2) a n d (β^{2} + 2) s u m = α^{2} + β^{2} + 4 = {(α + β)}^{2} - 2 α β + 4 = \frac{9}{4} + 6 + 4 = \frac{49}{4} p r o d u c t = (α^{2} + 2) (β^{2} + 2) = α^{2} β^{2} + 2 ({α^{2} + β}^{2}) + 4 = 9 + 4 + 2 [{(α + β)}^{2} - 2 α β] = 13 + \frac{33}{2} = \frac{59}{2} e q u a t i o n x^{2} - (s u m o f r o o t s) x + (p r o d u c t o f r o o t s) = 0 x^{2} - \frac{49}{4} x + \frac{59}{2} = 0 4 x^{2} - 49 x + 118 = 0$

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