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

${\mathrm{x}}^2+{60}^{1/4}\mathrm{x}+\mathrm{a}=0\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=-{60}^{1/4},\mathrm{\alpha \beta }=\mathrm{a}$

${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+2\mathrm{\alpha \beta }=\sqrt{60}$

${\mathrm{\alpha }}^2+{\beta }^2=\sqrt{60}-2\mathrm{a}$

${\mathrm{\alpha }}^4+{\mathrm{\beta }}^4+2{\mathrm{\alpha }}^2{\mathrm{\beta }}^2=60+4a^2-4\sqrt{60}\mathrm{a}$

$\Rightarrow -30+2{\mathrm{a}}^2=60+4{\mathrm{a}}^2-4\sqrt{60}\mathrm{a}$

$\begin{array}{l}\Rightarrow 2{\mathrm{a}}^2-4\sqrt{60}\mathrm{a}+90=0\end{array}$
Product of rots = 90/2 = 45