If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 1 = 0$,then what is the equation whose roots are $\frac{1}{\alpha - 2}$ and $\frac{1}{\beta - 2}$?

For which real value of $a$ do the roots of the quadratic equation $2x^2 - (a^3 + 8a - 1) x + a^2 - 4a = 0$ have opposite signs?

If $p$ and $q$ are distinct prime numbers and the equation $x^2 - px + q = 0$ has positive integers as its roots,then the roots of the equation are:

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then $x^4 + y^4 =$

Let $\alpha, \beta$ be two roots of the quadratic equation $x^2+ax-b=0, b \neq 0$. If the straight line $x \cos \theta + y \sin \theta = c$ touches the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ at the point $(\alpha, \beta)$, then $(\frac{a}{b})^2 + \frac{2}{b} =$

If each root of the equation $2x^3 + ax^2 - 8x + b = 0$ is reduced by $1$,then in the transformed equation thus formed,the term containing $x^2$ and the constant term vanish. The roots of the original equation are