The equation whose roots are squares of the roots of x⁴ + x³ + 2x² + x + 1 = 0 is

$$\begin{gathered} y=x^2⟹x=\sqrt{y} \\ replace x by \sqrt{y} in the given equstion \end{gathered}$$

$\begin{array}{r}⟹(\sqrt{\mathrm{y}}{)}^4+(\sqrt{\mathrm{y}}{)}^3+2(\sqrt{\mathrm{y}}{)}^2+\sqrt{\mathrm{y}}+1=0\\ ⟹{\mathrm{y}}^2+\mathrm{y}\sqrt{\mathrm{y}}+2\mathrm{y}+\sqrt{\mathrm{y}}+1=0\\ ⟹\sqrt{\mathrm{y}}(\mathrm{y}+1)=-\left({\mathrm{y}}^2+2\mathrm{y}+1\right)\\ ⟹\mathrm{y}(\mathrm{y}+1{)}^2=\left({\mathrm{y}}^2+2\mathrm{y}+1{)}^2\right.\end{array}$

$\begin{array}{r}⟹y\left(y^2+2y+1\right)=\left(y^4+4y^3+6y^2+4y+1\right)\\ ⟹y^3+2y^2+y=y^4+4y^3+6y^2+4y+1\\ ⟹y^4+3y^3+4y^2+3y+1=0\end{array}$