# Simplify $\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$ to $\frac{5+ \sqrt{5}}{8}$

I would like to know how the following expression was simplified?

From this $$\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$$

to this $$\frac{5+ \sqrt{5}}{8}$$

• Multiply the numerator and denominator by $6-\sqrt{20}$ and simplify Dec 1, 2012 at 20:26
• Thank you. I totally forgot this trick! Dec 4, 2012 at 3:27

You can multiply by the "conjugate": $$\frac{(5 +\sqrt{20})}{(6+\sqrt{20})}\frac{(6 - \sqrt{20})}{(6 - \sqrt{20})} = \dots$$
You could simplify it a step prematurely for kicks and giggles. $\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}$. Now, $\frac{5+\sqrt{20}}{6+\sqrt{20}}=\frac{5+2\sqrt{5}}{6+2\sqrt{5}}$. Then, of course, multiply by the fraction by $\frac{6-2\sqrt{5}}{6-2\sqrt{5}}$.
Multiplying by the conjugate to remove the radical is a common practice. It comes from the fact that $(a-b)(a+b)=a^2+ab-ba-b^2=a^2-b^2.$ So, in the case of $\frac{\text{something}}{u+\sqrt{v}}$, we multiply by $\frac{u-\sqrt{v}}{u-\sqrt{v}}$ since we know that $(u+\sqrt{v})(u-\sqrt{v})=u^2-v^2$. Similarly, if it is $u-\sqrt{v}$ in the denominator, we simply multiply by $\frac{u+\sqrt{v}}{u+\sqrt{v}}$.