If $\mathrm{\alpha },\mathrm{\beta }$ are the roots of the equation ${\mathrm{x}}^2-3\mathrm{x}+5=0$ and $\mathrm{\gamma },\mathrm{\delta }$ are the roots of the equation ${\mathrm{x}}^2+5\mathrm{x}-3=0$, then the equation whose roots are $\mathrm{\alpha \gamma }+\mathrm{\beta \delta } \text{and }\mathrm{\alpha \delta }+\mathrm{\beta \gamma }$ is

Solution:

$\because \mathrm{\alpha }+\mathrm{\beta }=3,\mathrm{\alpha \beta }=5,\mathrm{\gamma }+\mathrm{\delta }=(-5),\mathrm{\gamma \delta }=(-3)$

Sum of roots

$\begin{array}{rl} & =(\mathrm{\alpha \gamma }+\mathrm{\beta \delta })+(\mathrm{\alpha \delta }+\mathrm{\beta \gamma })\\ & =(\mathrm{\alpha }+\mathrm{\beta })(\mathrm{\gamma }+\mathrm{\delta })=3\times (-5)=(-15)\end{array}$

Product of roots

$\begin{array}{rl} & =(\mathrm{\alpha \gamma }+\mathrm{\beta \delta })(\mathrm{\alpha \delta }+\mathrm{\beta \gamma })\\ & ={\mathrm{\alpha }}^2\mathrm{\gamma \delta }+{\mathrm{\alpha \beta \gamma }}^2+{\mathrm{\beta \alpha \delta }}^2+{\mathrm{\beta }}^2\mathrm{\gamma \delta }\\ & =\mathrm{\gamma \delta }({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2)+\mathrm{\alpha \beta }({\mathrm{\gamma }}^2+{\mathrm{\delta }}^2)\end{array}$

$\begin{array}{rl} & =-3({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2)+5({\mathrm{\gamma }}^2+{\mathrm{\delta }}^2)\\ & =-3\left[\right(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }]+5\left[\right(\mathrm{\gamma }+\mathrm{\delta }{)}^2-2\mathrm{\gamma \delta }]\\ & =-3[9-10]+5[25+6]=158\end{array}$

Required equation is ${\mathrm{x}}^2+15\mathrm{x}+158=0$