How do you rationalize the denominator and simplify 12/(sqrt5-1)?

2 Answers
May 28, 2018

$3 \left(\sqrt{5} + 1\right)$

Explanation:

Use the identities

• $\frac{a}{a} = 1$ for every non-zero $a$
• $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

to write

$\setminus \frac{12}{\sqrt{5} - 1} = \setminus \frac{12}{\sqrt{5} - 1} \setminus \cdot 1 = \setminus \frac{12}{\sqrt{5} - 1} = \setminus \frac{12}{\sqrt{5} - 1} \setminus \cdot \setminus \frac{\sqrt{5} + 1}{\sqrt{5} + 1}$

Now, as you can see, the numerator is $12 \left(\sqrt{5} + 1\right)$, which is fine because we are allowed to have roots in the numerator.

The denominator, instead, is written in the same form as the second property written at the beginning:

$\left(\sqrt{5} + 1\right) \left(\sqrt{5} - 1\right) = {\left(\sqrt{5}\right)}^{2} + {1}^{2} = 5 - 1 = 4$

So, the fraction becomes

$\setminus \frac{12 \left(\sqrt{5} + 1\right)}{4} = 3 \left(\sqrt{5} + 1\right)$

May 28, 2018

$3 \sqrt{5} + 3$

Explanation:

When having a root on the bottom with an extra $+$ or $-$ on the bottom, we multiply by the opposite:

$\Rightarrow \frac{12}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1}$

$\Rightarrow \frac{12 \sqrt{5} + 12}{4}$

$\Rightarrow 3 \sqrt{5} + 3$