Solve the given two equations and select the correct option.
$I.$ $\sqrt{500} x + \sqrt{402} = 0$
$II.$ $\sqrt{360} y + \sqrt{200} = 0$

If $x$ is a solution of the equation $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$ where $x \ge \frac{1}{2}$,then $\sqrt{4x^2 - 1}$ is equal to:

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+px+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2x^{2}+2qx+1=0$,then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to:

If $(x^{3}-y^{3}):(x^{2}+xy+y^{2})=5:1$ and $(x^{2}-y^{2}):(x-y)=7:1$,then the ratio $2x:3y$ equals

What is the sum of the squares of the roots of the quadratic equation $x^2 - 3x + 1 = 0$?

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \ne 0$; $a, b, c$ being distinct),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = $