The equation whose roots are the square of the roots of the equation $2{\mathrm{x}}^2+3\mathrm{x}+1=0$

Solution:

Let the roots of equation
$2{\mathrm{x}}^2+3\mathrm{x}+1=0\text{ is }\mathrm{\alpha }\text{ and }\mathrm{\beta }$
Then ,$\mathrm{\alpha }+\mathrm{\beta }=-\frac{3}{2}\text{ and }\mathrm{\alpha \beta }=\frac{1}{2}$
$\begin{array}{r}\therefore {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }\\ ={\left(\frac{-3}{2}\right)}^2-2\times \frac{1}{2}=\frac{9}{4}-1=\frac{5}{4}\end{array}$
and ${\mathrm{\alpha }}^2{\mathrm{\beta }}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
$\therefore$ Required equation is
$\begin{array}{l}{\mathrm{x}}^2-\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)\mathrm{x}+{\mathrm{\alpha }}^2{\mathrm{\beta }}^2=0\\ \Rightarrow {\mathrm{x}}^2-\frac{5}{4}\mathrm{x}+\frac{1}{4}=0\\ \Rightarrow 4{\mathrm{x}}^2-5\mathrm{x}+1=0\end{array}$