Question

Let $a \in R$ and let $\alpha, \beta$ be the roots of the equation $x^2+60^{\frac{1}{4}} x+a=0$ If $\alpha^4+\beta^4=-30$, then the product of all possible values of $a$ is_____

Answer 1

The correct answer is 45.

α+β=−6041​&αβ=a
Given α4+β4=−30
⇒(α2+β2)2−2α2β2=−30
⇒{(α+β)2−2αβ}2−2a2=−30
⇒{6021​−2a}2−2a2=−30
⇒60+4a2−4a×6021​−2a2=−30
⇒2a2−4.6021​a+90=0
Product =290​=45