If $x = \sqrt{7 + 4\sqrt{3}},$ then $x + \frac{1}{x} = $

Solve the given two equations and provide the correct answer from the given options.
$I.$ $\frac{x}{x+7} + \frac{x+7}{x} = 12$
$II.$ $\frac{y}{y+8} + \frac{y+8}{y} = 16$

If $|{x^2} - x - 6| = x + 2$,then the values of $x$ are

Let $\alpha$ and $\beta$ be the roots of the equation $5x^{2} + 6x - 2 = 0$. If $S_{n} = \alpha^{n} + \beta^{n}$ for $n = 1, 2, 3, \ldots$,then:

If $k \in ( - \infty , - 2) \cup (2, \infty ),$ then the roots of the equation $x^2 + 2kx + 4 = 0$ are

If the roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$,then $p$ is equal to: