If $\left\{\frac{1}{2}(a-b)\right\}^{2}+a b=p(a+b)^{2},$ then the value of $p$ is (assume that $a \neq-b).$

If the sum of the roots of a quadratic equation is $1$ and the product of the roots is $-20$, find the quadratic equation.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + qx - r = 0$,then find the equation whose roots are $\left( \beta \gamma + \frac{1}{\alpha} \right), \left( \gamma \alpha + \frac{1}{\beta} \right), \left( \alpha \beta + \frac{1}{\gamma} \right)$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x + 3y = 77$
$II.$ $2x + 5y = (2601)^{1/2}$

Let $p, q$ and $r$ be real numbers $(p \ne q, r \ne 0),$ such that the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign. Then the sum of squares of these roots is equal to:

If a root of the equation $x^2 + px + 12 = 0$ is $4$,while the roots of the equation $x^2 + px + q = 0$ are equal,then the value of $q$ will be