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

$\alpha+\beta=-60^{\frac{1}{4}} \& \alpha \beta= a$
Given $\alpha^4+\beta^4=-30$
$\Rightarrow\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=-30$
$ \Rightarrow\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}^2-2 a^2=-30$
$ \Rightarrow\left\{60^{\frac{1}{2}}-2 a \right\}^2-2 a ^2=-30$
$ \Rightarrow 60+4 a ^2-4 a \times 60^{\frac{1}{2}}-2 a ^2=-30$
$\Rightarrow 2 a ^2-4.60^{\frac{1}{2}} a +90=0$
Product $=\frac{90}{2}=45$