# How do you simplify 5/(sqrt3 - 2)?

Apr 6, 2015

When we simplify radicals in a fraction, we have to rationalize the denominator where the radical is at.

There is a rule that shows $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$.
So in the fraction, $\frac{5}{\sqrt{3} - 2}$ , the denominator $\left(\sqrt{3} - 2\right)$ represents (a-b) in the rule of $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$. To get rid of a surd, we can square the surd as seen in ${a}^{2} - {b}^{2}$

Let's place $\sqrt{3}$ as $a$ and $2$ as $b$.

Hence, $\left(\sqrt{3} - 2\right)$$\left(\sqrt{3} + 2\right)$ = ${\left(\sqrt{3}\right)}^{2}$- ${2}^{2}$

As we remember, the square of a radical/surd rationalises the radical.
${\left(\sqrt{3}\right)}^{2}$= $3$

So what we can do is to muiltiply $\frac{5}{\sqrt{3} - 2}$ x $\frac{\sqrt{3} + 2}{\sqrt{3} + 2}$ where $\frac{\sqrt{3} + 2}{\sqrt{3} + 2}$= $1$ so it does not make a difference to the value of the product.

(5(sqrt3+2))/((sqrt3 - 2)(sqrt3+2)=$\frac{5 \sqrt{3} + 10}{{\left(\sqrt{3}\right)}^{2} - {2}^{2}}$=$\frac{5 \sqrt{3} + 10}{3 - 4}$=$\frac{5 \sqrt{3} + 10}{- 1}$= $- 5 \sqrt{3} - 10$

$\frac{5}{\sqrt{3} - 2}$=$- 5 \sqrt{3} - 10$=$- 18.66 \left(2 d . p .\right)$