Question

Rationalize the denominator and simplify.

$\frac{\sqrt{x}-2}{\sqrt{x}+2}$

Short answer

Rationalized form of the given expression is $\frac{{\left(\sqrt{x}-2\right)}^2}{\left(x-4\right)}$.

Step-by-step solution

Step 1. To determine the conjugate if present. Otherwise simply eliminate the root terms using multiplication on numerator and denominator.

$\frac{\sqrt{x}-2}{\sqrt{x}+2}\times \frac{\sqrt{x}-2}{\sqrt{x}-2}$

Step 2. Re write the expression after multiplying.

$\frac{\sqrt{x}-2}{\sqrt{x}+2}\times \frac{\sqrt{x}-2}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}$

Step 3. Simplify the expression.

$\begin{array}{c}\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{{\left(\sqrt{x}-2\right)}^2}{\left({\left(\sqrt{x}\right)}^2-2^2\right)}\\ =\frac{{\left(\sqrt{x}-2\right)}^2}{\left(x-4\right)}\end{array}$