Question

Choose and write the correct option. On rationalization of the denominator of the number $\frac{\sqrt{5}+2}{\sqrt{5}-2}$ we get:

• A. ${\left(\sqrt{5}+2\right)}^2$
• B. $\left(\sqrt{5}\right)2$
• C. $3$
• D. None of these

Answer 1

The correct option is A

${\left(\sqrt{5}+2\right)}^2$

The explanation for the correct option:

• "When the denominator of an expression contains a term with a square root (or a number under a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called rationalizing the denominator."
• To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator to remove the radical sign from the denominator.

Given expression: $\frac{\sqrt{5}+2}{\sqrt{5}-2}$

Multiply the numerator and denominator by $\sqrt{5}+2$ (i.e., the conjugate of $\sqrt{5}-2$):

$\begin{array}{rcl}\frac{\sqrt{5}+2}{\sqrt{5}-2}& =& \frac{\sqrt{5}+2}{\sqrt{5}-2}\times \frac{\sqrt{5}+2}{\sqrt{5}+2}\\ & =& \frac{{\left(\sqrt{5}+2\right)}^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\end{array}$

Use identity $\left(a+b\right)\left(a-b\right)\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}$ in the denominator:

$\begin{array}{rcl}\frac{\sqrt{5}+2}{\sqrt{5}-2}& =& \frac{{\left(\sqrt{5}+2\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(2\right)}^2}\\ & =& \frac{{\left(\sqrt{5}+2\right)}^2}{5-4}\\ & =& \frac{{\left(\sqrt{5}+2\right)}^2}{1}\\ & =& {\left(\sqrt{5}+2\right)}^2\end{array}$

Therefore, on rationalizing the denominator of the number $\frac{\sqrt{5}+2}{\sqrt{5}-2}$, we get $\begin{array}{rcl}& & {\left(\sqrt{5}+2\right)}^2\end{array}$.

Hence, the correct option is A.