Question

Rationalize the denominator$\frac{16}{\sqrt{41}-5}$.

Answer 1

Rationalization is a process by which radicals in the denominator of an algebraic fraction are eliminated.
Multiply numerator & denominator by $\left(\sqrt{41}+5\right)$ we have:

$$\begin{gathered} \frac{16}{\sqrt{41}-5}=\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5} \\ =\frac{16\left(\sqrt{41}+5\right)}{{\left(\sqrt{41}\right)}^2-5^2}[\left(a+b\right)\left(a-b\right)=a^2-b^2] \\ =\frac{16\left(\sqrt{41}+5\right)}{41-25} \\ =\frac{16(\sqrt{41}+5)}{16} \\ =\sqrt{41}+5 \end{gathered}$$

So, after rationalization of denominator the fraction$\frac{16}{\sqrt{41}-5}$ becomes $\sqrt{41}+5$.