Question

Rationalize the numerator.

$\frac{\sqrt{3}+\sqrt{5}}{2}$

Short answer

The solution is$\frac{1}{\left(\sqrt{5}-\sqrt{3}\right)}$

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

$\frac{\sqrt{3}+\sqrt{5}}{2}$

Here the numerator is $\sqrt{3}+\sqrt{5}$, so multiply the numerator and denominator by$\sqrt{3}-\sqrt{5}$

Step 4. Rationalize the numerator.

Consider$\frac{\sqrt{3}+\sqrt{5}}{2}$

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

role="math" localid="1643279385012" $\begin{array}{l}\frac{\sqrt{3}+\sqrt{5}}{2}=\frac{\sqrt{3}+\sqrt{5}}{2}\cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}{2\left(\sqrt{3}-\sqrt{5}\right)}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{5}\right)}^2}{2\left(\sqrt{3}-\sqrt{5}\right)}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{3-5}{2\left(\sqrt{3}-\sqrt{5}\right)}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{-2}{2\left(\sqrt{3}-\sqrt{5}\right)}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{-1}{\left(\sqrt{3}-\sqrt{5}\right)}\\ \Rightarrow \frac{\sqrt{3}+\sqrt{5}}{2}=\frac{1}{\left(\sqrt{5}-\sqrt{3}\right)}\end{array}$