Question

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

Answer 1

`3(sqrt(5)+1)`

Explanation:

Use the identities

• `a/a=1` for every non-zero `a`
• `(a+b)(a-b) = a^2-b^2`

to write

`\frac{12}{sqrt(5)-1} = \frac{12}{sqrt(5)-1} \cdot 1 = \frac{12}{sqrt(5)-1} = \frac{12}{sqrt(5)-1}\cdot \frac{sqrt(5)+1}{sqrt(5)+1}`

Now, as you can see, the numerator is `12(sqrt(5)+1)`, 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:

`(sqrt(5)+1)(sqrt(5)-1) = (sqrt(5))^2 + 1^2 = 5-1=4`

So, the fraction becomes

`\frac{12(sqrt(5)+1)}{4} = 3(sqrt(5)+1)`

Answer 2

`3sqrt5+3`

Explanation:

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

`rArr (12)/(sqrt5-1) xx (sqrt5+1)/(sqrt5+1)`

`rArr (12sqrt5+12)/4`

`rArr 3sqrt5+3`