Question

Rationalize the denominator.

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

Short answer

The solution is$-3\left(2+\sqrt{5}\right)$

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

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

Here the denominator is $2-\sqrt{5}$, so multiply the numerator and denominator by$2+\sqrt{5}$

Step 4. Rationalize the denominator.

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

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

role="math" localid="1643225460591" $\begin{array}{l}\frac{3}{2-\sqrt{5}}=\frac{3}{2-\sqrt{5}}\cdot \frac{2+\sqrt{5}}{2+\sqrt{5}}\\ \Rightarrow \frac{3}{2-\sqrt{5}}=\frac{3\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\\ \Rightarrow \frac{3}{2-\sqrt{5}}=\frac{3\left(2+\sqrt{5}\right)}{2^2-{\left(\sqrt{5}\right)}^2}\\ \Rightarrow \frac{3}{2-\sqrt{5}}=\frac{3\left(2+\sqrt{5}\right)}{4-5}\\ \Rightarrow \frac{3}{2-\sqrt{5}}=-3\left(2+\sqrt{5}\right)\end{array}$