$\frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}=?$

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:

Let one root of $ax^2 + bx + c = 0$,where $a, b, c$ are integers,be $3 + \sqrt{5}$. Then the other root is:

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

Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$,$\beta$ is a root of $a^2x^2 - bx - c = 0$,and $0 < \alpha < \beta$,then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies:

If both the roots of the quadratic equation ${x^2} - 2kx + {k^2} + k - 5 = 0$ are less than $5$,then $k$ lies in the interval