Question

Rationalize the numerator.

$\sqrt{x^2+1}-x$

Short answer

The solution is$\frac{1}{\sqrt{x^2+1}+x}$

Step-by-step solution

Step 1. Apply the concept of rationalizing the numerator.

If a fraction has a numerator of the form$A+B\sqrt{C}$, we can rationalize the numerator 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

$\sqrt{x^2+1}-x$

Here the numerator is $\sqrt{x^2+1}-x$, so multiply the numerator and denominator by$\sqrt{x^2+1}+x$

Step 4. Rationalize the numerator.

Consider$\sqrt{x^2+1}-x$

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

role="math" localid="1643282048337" $\begin{array}{l}\sqrt{x^2+1}-x=\sqrt{x^2+1}-x\cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\\ \Rightarrow \sqrt{x^2+1}-x=\frac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}\\ \Rightarrow \sqrt{x^2+1}-x=\frac{{\left(\sqrt{x^2+1}\right)}^2-{\left(x\right)}^2}{\sqrt{x^2+1}+x}\\ \Rightarrow \sqrt{x^2+1}-x=\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}\\ \Rightarrow \sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+x}\end{array}$