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

Answer: 45

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