If $\mathrm{\alpha }$ and $\mathrm{\beta }$ are the roots of equation 2x² - 5x + 7 = 0, then the equation whose roots are 2$\mathrm{\alpha }$ + 3$\mathrm{\beta }$ and 3$\mathrm{\beta }$ + 2$\mathrm{\beta }$, is

Since, $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of the equation
2x² -5x+ 7 =0,then
$\mathrm{\alpha }+\mathrm{\beta }=\frac{5}{2}\text{ and }\mathrm{\alpha \beta }=\frac{7}{2}$
Now sum of roots = $(2\mathrm{\alpha }+3\mathrm{\beta })+(3\mathrm{\alpha }+2\mathrm{\beta })=5(\mathrm{\alpha }+\mathrm{\beta })=\left(\frac{25}{2}\right)$
and product of roots 
 $=(2\mathrm{\alpha }+3\mathrm{\beta })(3\mathrm{\alpha }+2\mathrm{\beta })$
$=6\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)+13\mathrm{\alpha \beta }$
$\begin{array}{l}=6\left[(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }\right]+13\mathrm{\alpha \beta }\\ =\left[6\times \left(\frac{25}{4}-7\right)+\frac{91}{2}\right]=41\end{array}$
The required equation is
$\begin{array}{l}{\mathrm{x}}^2-\frac{25}{2}\mathrm{x}+41=0\\ \Rightarrow 2{\mathrm{x}}^2-25\mathrm{x}+82=0\end{array}$