Solve the given two equations and select the correct answer from the given options.
$I.$ $(17)^{2} + 144 \div 18 = x$
$II.$ $(26)^{2} - 18 \times 21 = y$

Solution of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

$I. x^{2}+10 x+24=0$
$II. 4 y^{2}-17 y+18=0$

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

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4sqrt{ab}x + 1 = 0$ will be