Let  $a,b,c$ be real numbers  $a\ne 0$. If $\alpha$ is a root  $a^2{x}^2+bx+c=0$,  $\beta$ is a root of  $a^2{x}^2-bx-c=0$ and $0<\alpha <\beta$, then the equation $a^2{x}^2+2bx+2c=0$ has a root $\gamma$ that always satisfies

Since $\alpha \text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}\beta$ are the roots of given equations. So we have $a^2{\alpha }^2+b\alpha +c=0$ and $a^2{\beta }^2-b\beta -c=0$.

Let $f\left(x\right)=a^2{x}^2+2bx+2c=0$

Then $f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=0$

$=a^2{\alpha }^2+2(b\alpha +c)=a^2{\alpha }^2-2a^2{\alpha }^2=-a^2{\alpha }^2=-ve$

and $f\left(\beta \right)=a^2{\beta }^2+2(b\beta +c)=a^2{\beta }^2+2a^2{\beta }^2$ $=3a^2{\beta }^2=+ve$

Since $f\left(\alpha \right)andf\left(\beta \right)$ are of opposite  signs, therefore by theory of equations there lies a root $\gamma$ of the equation $f\left(x\right)=0$ between $\alpha$ and $\beta$ i.e. $\alpha <\gamma <\beta$