If $\alpha , \beta$ are the nonzero roots of ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ and ${\mathrm{\alpha }}^2, {\mathrm{\beta }}^2$ are the roots of ${\mathrm{a}}^2{\mathrm{x}}^2+{\mathrm{b}}^2\mathrm{x}+{\mathrm{c}}^2=0$, then a, b, c are in

 $\mathrm{\alpha }, \mathrm{\beta }$ are roots of ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$. Hence,

$\begin{array}{l}\mathrm{\alpha }+\mathrm{\beta }=-\frac{\mathrm{b}}{\mathrm{a}} \left(1\right)\\ \mathrm{\alpha \beta }=\frac{\mathrm{c}}{\mathrm{a}} \left(2\right)\end{array}$

${\mathrm{\alpha }}^2, {\mathrm{\beta }}^2$are roots of ${\mathrm{a}}^2{\mathrm{x}}^2+{\mathrm{b}}^2\mathrm{x}+{\mathrm{c}}^2=0$. Hence,

$\begin{array}{l}{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2} \left(3\right)\\ {\mathrm{\alpha }}^2{\mathrm{\beta }}^2=\frac{{\mathrm{c}}^2}{{\mathrm{a}}^2} \left(4\right)\end{array}$

Now, from (3), we have

$(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }=-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$

$\Rightarrow {\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)}^2-2\frac{\mathrm{c}}{\mathrm{a}}=\frac{-{\mathrm{b}}^2}{{\mathrm{a}}^2}$

or $2\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{2\mathrm{c}}{\mathrm{a}}$

or ${\mathrm{b}}^2=\mathrm{ac}\Rightarrow \mathrm{a},\mathrm{b},\mathrm{c}$ are in G.P.