If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

Solve the given two equations and select the correct option.
$I.$ $9x^{2} - 114x + 361 = 0$
$II.$ $y^{2} = 36$

Let $\alpha, \beta$ be the roots of $x^2 - x + p = 0$ and $\gamma, \delta$ be the roots of $x^2 - 4x + q = 0$. If $\alpha, \beta, \gamma, \delta$ are in $G.P.$,then the integral values of $p, q$ are respectively:

Let $x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ and $y = \frac{1}{x}$. Then the value of $3x^2 - 5xy + 3y^2$ is:

Suppose that $x$ and $y$ are positive numbers with $xy = \frac{1}{9}$,$x(y + 1) = \frac{7}{9}$,and $y(x + 1) = \frac{5}{18}$. The value of $(x + 1)(y + 1)$ equals:

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - a(x + 1) - b = 0$,then $(\alpha + 1)(\beta + 1) = $