Let and let be the roots of the equation If , then the product of all possible values of is __________ . [2023]

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Solution

Q.
Let $a \in R$ and let $α, β$ be the roots of the equation $x^{2} + 60^{\frac{1}{4}} x + a = 0 .$

If $α^{4} + β^{4} = - 30$ , then the product of all possible values of $a$ is __________ . [2023]

Ans.
(45)

$x^{2} + 60^{\frac{1}{4}} x + a = 0 ... (i)$

As we have, $α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$

$= ({{(α + β)}^{2} - 2 α β)}^{2} - 2 α^{2} β^{2} ... (ii)$

Now, $α + β = - 60^{1 / 4} and α β = a [From (i)]$

${(α + β)}^{2} = {(60)}^{\frac{1}{2}} = \sqrt{60} and α^{2} \cdot β^{2} = a^{2} ... (iii)$

Also, $α^{4} + β^{4} = - 30 ... (iv)$

So, substituting values from (iii) and (iv) in (ii), we get:

$a^{2} - 2 \sqrt{60} a + 45 = 0$

$\Rightarrow a^{2} - 4 \sqrt{15} a + 45 = 0 \Rightarrow (a - 3 \sqrt{15}) (a - \sqrt{15}) = 0$

Product of values of $a$ = $3 \sqrt{15} \times \sqrt{15} = 3 \times 15 = 45$

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