To find the value of \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\)
Take the L.C.M
\(\Rightarrow \frac{\beta^2+\alpha^2}{\alpha^2 \beta^2}=\frac{\beta^2+a^2}{(\alpha \beta)^2}\)
Where \(\alpha^2+\beta^2 \equiv(\alpha+\beta)^2-2 \alpha \beta\)
Since \(\alpha\) and \(\beta\) are the roots of the equation
\begin{array}{l}
x^2-10 x+2=0 \\
a=1, b=-10, c=2
\end{array}
then sum of the roots
\begin{array}{l}
\alpha+\beta=-\frac{b}{a} \\
\alpha+\beta=-\frac{(-10)}{1}=10 \\
\alpha \beta=\frac{c}{a}=\frac{2}{1}=2
\end{array}
thus
\begin{aligned}
\frac{\alpha^2+\beta^2}{(\alpha \beta)^2} &=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2} \\
&=\frac{10^2-2(2)}{(2)^2} \\
&=\frac{100-4}{4}=\frac{96}{4}=24
\end{aligned}