$\begin{align}
& \text{If the roots of the equation }{{x}^{2}}+2{{x}^{2}}-1=0\text{ are }\alpha \text{ and }\beta \text{, } \\
& \text{form the equation whose roots are (i) }{{\alpha }^{2}},{{\beta }^{2}}\text{ (ii) }{{\alpha }^{2}}+{{\beta }^{2}},4\alpha \beta \\
\end{align}$
$\begin{align}
& \text{The given equation }{{x}^{2}}+2x-1=0, \\
& a=1,b=2,c=-1 \\
& \text{If }\alpha \text{ and }\beta \text{ are roots of the equation therefore } \\
& \text{Sum of roots }=\alpha +\beta =-\tfrac{b}{a}=-\tfrac{2}{1}=-2 \\
& \text{Product of roots}=\alpha \beta =\tfrac{c}{a}=-1 \\
& \text{Sum of roots for }{{\alpha }^{2}}\text{ and }{{\beta }^{2}}:{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2(\alpha \beta )={{(-2)}^{2}}-2(-1)=4+2=6 \\
& \text{Product of roots for }{{\alpha }^{2}}\text{ and }{{\beta }^{2}}={{\alpha }^{2}}\times {{\beta }^{2}}={{(\alpha \beta )}^{2}}={{(-1)}^{2}}=1 \\
& \text{The new equation will be: }{{x}^{2}}-({{\alpha }^{2}}+{{\beta }^{2}})x+({{\alpha }^{2}}{{\beta }^{2}})={{x}^{2}}-6x+1=0 \\
& \text{The equation }{{x}^{2}}-6x+1=0\text{ has }{{\alpha }^{2}}\text{ and }{{\beta }^{2}}\text{ as roots} \\
& \text{Sum of roots for }{{\alpha }^{2}}+{{\beta }^{2}}\text{ and }4\alpha \beta \\
& ({{\alpha }^{2}}+{{\beta }^{2}})+4\alpha \beta =[{{(\alpha +\beta )}^{2}}-2\alpha \beta ]+4\alpha \beta ={{(\alpha +\beta )}^{2}}+2\alpha \beta \\
& ({{\alpha }^{2}}+{{\beta }^{2}})+4\alpha \beta ={{(-2)}^{2}}+2(-1)=4-2=2 \\
& \text{Product of roots for }{{\alpha }^{2}}+{{\beta }^{2}}\text{ and }4\alpha \beta \\
& ({{\alpha }^{2}}+{{\beta }^{2}}\text{)}4\alpha \beta =[{{(\alpha +\beta )}^{2}}-2\alpha \beta ]4\alpha \beta \\
& [{{(\alpha +\beta )}^{2}}-2\alpha \beta ]4\alpha \beta =\left[ {{(-2)}^{2}}-2(-1) \right]4(-1)=(4+2)(-4)=-24 \\
& \text{The new equation will be:}={{x}^{2}}-2x-24=0 \\
\end{align}$