If $x+\sqrt{15}=4,$ then find the value of $x+\frac{1}{x}$.

The correct option is B: $8$

$x+\sqrt{15}=4$

$\Rightarrow x=4-\sqrt{15}$

and,

$\frac{1}{x}=\frac{1}{4-\sqrt{15}}$

$=\frac{1\times (4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}$ (Rationalising the denominator)

$=\frac{4+\sqrt{15}}{(4{)}^2-(\sqrt{15}{)}^2}$ [$\because a^2-b^2=(a-b)(a+b)$]

$=\frac{4+\sqrt{15}}{16-15}$

$=\frac{4+\sqrt{15}}{1}$

$=4+\sqrt{15}$

Now, $x+\frac{1}{x}=(4-\sqrt{15})+(4+\sqrt{15})=8$