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 $\frac{1}{\alpha_1^2}+\frac{1}{\alpha_2^2}+\frac{1}{\alpha_3^2}+\frac{1}{\alpha_4^2}+\frac{1}{\alpha_5^2}=$

Let $\alpha, \beta, \gamma$ be the roots of $x^3+x+10=0$ and $\alpha_1=\frac{\alpha+\beta}{\gamma^2}, \beta_1=\frac{\beta+\gamma}{\alpha^2}, \gamma_1=\frac{\gamma+\alpha}{\beta^2}$. Then,the value of $(\alpha_1^3+\beta_1^3+\gamma_1^3)-\frac{1}{10}(\alpha_1^2+\beta_1^2+\gamma_1^2)$ is

The cubic equation whose roots are the squares of the roots of $x^3-2x^2+10x-8=0$ is

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

The harmonic mean of two numbers is $-\frac{8}{5}$ and their geometric mean is $2$. The quadratic equation whose roots are twice those numbers is

If $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$ and $\gamma$ is a root of $x^{2}+px+1=0,$ then $(\alpha+\gamma)(\beta+\gamma)$ is