If $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ are the roots of $x^5-5 x^4+9 x^3-9 x^2$ $+5 x-1=0$ then $\frac{1}{\alpha_1^2}+\frac{1}{\alpha_2^2}+\frac{1}{\alpha_3^2}+\frac{1}{\alpha_4^2}+\frac{1}{\alpha_5^2}=$

$x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$ ...(i) $\because a_0 x^{\mathrm{n}}+a_1 x^{\mathrm{n}-1}+a_2 x^{\mathrm{n}-2}+\ldots .+a_{\mathrm{n}-1} x+a_{\mathrm{n}}=0$
Here $a_0=-a_{\mathrm{n}}, a_1=-a_{\mathrm{n}-1}$ is class II reciprocal equation. So, eq (i) is an odd degree reciprocal equation By inspection $x=1$ is a root Using synthetic division \begin{array}{c|cccccc}1 & 1 & -5 & 9 & -9 & 5 & -1 \\& 0 & 1 & -4 & 5 & -4 & 1 \\\hline & 1 & -4 & 5 & -4 & 1 & 0\end{array} $x^4-4 x^3+5 x^2-4 x+1=0$ ...(ii) Equation (ii) is an even degree reciprocal equation of class I. $\frac{x^4-4 x^3+5 x^2-4 x+1}{x^2}=0 \Rightarrow\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+5=0$....(iii) Let $x+\frac{1}{x}=a$. So, Eq (iii) implies. $\begin{array}{ll} \left(a^2-2\right)-4 a+5=0 \Rightarrow(a-3)(a-1)=0 \\ \Rightarrow a=1 \text { or } a=3 & a=3 \\ \Rightarrow a=1 & x+\frac{1}{x}=3 \\ \Rightarrow x+\frac{1}{x}=1 & \Rightarrow x^2-3 x+1=0 \\ \Rightarrow x^2-x+1=0 & x=\frac{3 \pm \sqrt{5}}{2} \end{array}$
So, $\alpha_1=1, \alpha_2=\frac{1+i \sqrt{3}}{2}, \alpha_3=\frac{1-i \sqrt{3}}{2}$ $\begin{aligned} & \alpha_4=\frac{3+\sqrt{5}}{2}, \alpha_5=\frac{3-\sqrt{5}}{2} \\ & \frac{1}{\alpha_1^2}+\frac{1}{\alpha_2^2}+\frac{1}{\alpha_3^2}+\frac{1}{\alpha_4^2}+\frac{1}{\alpha_5^2}=1+\left(\frac{2}{1+i \sqrt{3}}\right)^2+ \\ & \qquad\left(\frac{2}{1-i \sqrt{3}}\right)^2+\left(\frac{2}{3+\sqrt{5}}\right)^2+\left(\frac{2}{3-\sqrt{5}}\right)^2 \\ & =1+\frac{(2-2 i \sqrt{3})^2}{16}+\frac{(2+2 i \sqrt{3})^2}{16}+\frac{(6-2 \sqrt{5})^2}{16}+\frac{(6+2 \sqrt{5})^2}{16} \\ & =1+\frac{4-12+4-12+36+20+36+20}{16}=1+6=7 \end{aligned}$