If $\alpha , \beta$ arc the roots of ${\mathrm{ax}}^2+\mathrm{c}=\mathrm{bx}$, then the equation $(\mathrm{a}+\mathrm{cy}{)}^2={\mathrm{b}}^2\mathrm{y}$ in y has the roots

$\begin{array}{l}{\mathrm{ax}}^2-\mathrm{bx}+\mathrm{c}=0\\ \mathrm{\alpha }+\mathrm{\beta }=\frac{\mathrm{b}}{\mathrm{a}},\mathrm{a\beta }=\frac{\mathrm{c}}{\mathrm{a}}\end{array}$

Also, $(\mathrm{a}+\mathrm{cy}{)}^2={\mathrm{b}}^2\mathrm{y}$

$\begin{array}{l}\Rightarrow {\mathrm{c}}^2{\mathrm{y}}^2-\left({\mathrm{b}}^2-2\mathrm{ac}\right)\mathrm{y}+{\mathrm{a}}^2=0\\ \Rightarrow {\left(\frac{\mathrm{c}}{\mathrm{a}}\right)}^2{\mathrm{y}}^2-\left({\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^2-2\left(\frac{\mathrm{c}}{\mathrm{a}}\right)\right)\mathrm{y}+1=0\end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{\alpha \beta }{)}^2{\mathrm{y}}^2-\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)\mathrm{y}+1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}^2-\left({\mathrm{\alpha }}^{-2}+{\mathrm{\beta }}^{-2}\right)\mathrm{y}+{\mathrm{\alpha }}^{-2}{\mathrm{\beta }}^{-2}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(\mathrm{y}-{\mathrm{\alpha }}^{-2}\right)\left(\mathrm{y}-{\mathrm{\beta }}^{-2}\right)=0\end{array}$

Hence, the roots are ${\mathrm{\alpha }}^{-2}, {\mathrm{\beta }}^{-2}$