Solve the given two equations and choose the correct answer from the given options.
$I.$ $x^{2}+12x+36=0$
$II.$ $y^{2}+15y+56=0$

The reciprocal of $x+\frac{1}{x}$ is

If $\left(\frac{p^{-1} q^{2}}{p^{3} q^{-2}}\right)^{\frac{1}{3}}+\left(\frac{p^{5} q^{-3}}{p^{-2} q^{3}}\right)^{\frac{1}{3}}=p^{a} q^{b},$ then the value of $a+b,$ where $p$ and $q$ are different positive positive primes,is

Let $y = f(x) = ax^2 + 2bx + c$ (where $a, b, c \in R$ and $a \neq 0$). If $f(x) = 0$ has imaginary roots and $4a + 4b + c < 0$,then which of the following is true?

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

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: