If $\mathrm{\alpha },\mathrm{\beta }$ are real and ${\mathrm{\alpha }}^2,{\mathrm{\beta }}^2$ are the roots of the equation ${\mathrm{a}}^2{\mathrm{x}}^2+\mathrm{x}+1-{\mathrm{a}}^2=0(\mathrm{a}>1),\text{ then }{\mathrm{\beta }}^2=$

$$\begin{gathered} {\alpha }^2,{\beta }^2\text{ are the roots of }{\mathrm{a}}^2{\mathrm{x}}^2+\mathrm{x}+1-{\mathrm{a}}^2=0 \\ \Rightarrow {\alpha }^2+{\beta }^2=-\frac{1}{{\mathrm{a}}^2},\text{ And } \\ \Rightarrow {\alpha }^2\left({\beta }^2\right)=\frac{1-{\mathrm{a}}^2}{{\mathrm{a}}^2} \\ {\left({\alpha }^2-{\beta }^2\right)}^2={\left({\alpha }^2+{\beta }^2\right)}^2-4{\alpha }^2{\beta }^2 \\ =\frac{1}{{\mathrm{a}}^4}+4\left(\frac{{\mathrm{a}}^2-1}{{\mathrm{a}}^2}\right)=\frac{1}{{\mathrm{a}}^4}+4-\frac{4}{{\mathrm{a}}^2} \\ ={\left(2-\frac{1}{{\mathrm{a}}^2}\right)}^2\Rightarrow {\alpha }^2-{\beta }^2=2-\frac{1}{{\mathrm{a}}^2} \\ {\beta }^2=\frac{1}{2}\left[\left({\alpha }^2+{\beta }^2\right)-\left({\alpha }^2-{\beta }^2\right)\right] \end{gathered}$$

$1-\frac{1}{{\mathrm{a}}^2}$