If the product of the roots of the equation $2x^2 + 6x + \alpha^2 + 1 = 0$ is $-\alpha$,then the value of $\alpha$ will be

The value of $a$ $(a \ge 3)$ for which the sum of the cubes of the roots of $x^2 - (a - 2)x + (a - 3) = 0$ assumes the least value is:

Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3}x-16=0$,and $\gamma$ and $\delta$ be the roots of $x^2+3x-1=0$. If $P_{n}=\alpha^{n}+\beta^{n}$ and $Q_{n}=\gamma^{n}+\delta^{n}$,then $\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to

Suppose that the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,both of which are different from $\frac{1}{3}$. Then,an equation whose roots are $\frac{1}{3\alpha-1}$ and $\frac{1}{3\beta-1}$ is

Let $\alpha$ and $\beta$ be the roots of the equation $p x^2 + q x + r = 0$,where $p \neq 0$. If $p, q, r$ are in $AP$ and $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,then the value of $|\alpha - \beta|$ is

If $\alpha, \beta$ are the roots of $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of $x^2 + qx + 1 = 0$,then $(\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta) = \dots$