Let a R and let be the roots of the equation x 2 + 60 1 4 x + a = 0 . If 4 + 4 = - 30 , then the product of all possible values of a is

Question

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 _______.

Smart Achievers · Accepted Answer

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

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

Also, $α^{4} + β^{4} = - 30$ ...(iv)

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

$a^{2} - 2 \sqrt{60} a + 45 = 0$ $\Rightarrow a^{2} - 4 \sqrt{15} a + 45 = 0$

$\Rightarrow (a - 3 \sqrt{15}) (a - \sqrt{15}) = 0$

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

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 _______.

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Solution

Q. Let a∈R and let α, β be the roots of the equation x2+6014x+a=0. If α4+β4=-30, then the product of all possible values of a is _______.

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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 _______.