$\begin{align} & \text{If the roots of the equation }2{{x}^{2}}+x+1=0\text{ are }\alpha \text{ and }\beta .\text{ Form the equation whose roots are } \\ & \tfrac{1}{{{\alpha }^{2}}}\text{ and }\tfrac{1}{{{\beta }^{2}}} \\\end{align}$
$\begin{align} & a=2,b=1,\text{ }c=1 \\ & \text{Sum of roots }\alpha +\beta =-\tfrac{b}{a}=-\tfrac{1}{2} \\ & \text{Product of roots }\alpha \beta =\tfrac{c}{a}=\tfrac{1}{2} \\ & \text{For the new equation whose roots are } \\ & \tfrac{1}{{{\alpha }^{2}}}\text{ and }\tfrac{1}{{{\beta }^{2}}} \\ & \text{Sum of new roots}=\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{(\alpha \beta )}^{2}}}=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \beta )}^{2}}} \\ & \text{Sum of new roots}=\frac{{{(-\tfrac{1}{2})}^{2}}-2(\tfrac{1}{2})}{(\tfrac{1}{2})}=\frac{\tfrac{1}{4}-1}{\tfrac{1}{4}}=\frac{-\tfrac{3}{4}}{\tfrac{1}{4}}=-3 \\ & \text{Product of new roots}=\frac{1}{{{\alpha }^{2}}}\times \frac{1}{{{\beta }^{2}}}=\frac{1}{{{(\alpha \beta )}^{2}}}=\frac{1}{{{(\tfrac{1}{2})}^{2}}}=4 \\ & \text{The new equation}={{x}^{2}}-(\text{sum of new roots)}x+(\text{product of new roots)}=0 \\ & ={{x}^{2}}+3x+4=0 \\\end{align}$