Question

Rationalize the denominator.

$\frac{1}{\sqrt{x}+1}$

Short answer

The solution is$\frac{\sqrt{x}-1}{x-1}$

Step-by-step solution

Step 1. Apply the concept of rationalizing the denominator.

If a fraction has a denominator of the form$A+B\sqrt{C}$, we can rationalize the denominator by multiplying numerator and denominator by the conjugate radical $A-B\sqrt{C}$.

Step 2. Apply the concept of special multiplication.

If A and B are real numbers, then

$\left(A-B\right)\left(A+B\right)=A^2-B^2$

Step 3. Analyze the expression.

Given

$\frac{1}{\sqrt{x}+1}$

Here the denominator is $\sqrt{x}+1$, so multiply the numerator and denominator by$\sqrt{x}-1$

Step 4. Rationalize the denominator.

Consider$\frac{1}{\sqrt{x}+1}$

So, the simplified expression after rationalizing denominator is given by:

role="math" localid="1643227648059" $\begin{array}{l}\frac{1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}+1}\cdot \frac{\sqrt{x}-1}{\sqrt{x}-1}\\ \Rightarrow \frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ \Rightarrow \frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{{\left(\sqrt{x}\right)}^2-1^2}\\ \Rightarrow \frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{x-1}\end{array}$