How do you rationalize the denominator and simplify 12/(sqrt5-1)1251?

2 Answers
May 28, 2018

3(sqrt(5)+1)3(5+1)

Explanation:

Use the identities

  • a/a=1aa=1 for every non-zero aa
  • (a+b)(a-b) = a^2-b^2(a+b)(ab)=a2b2

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}1251=12511=1251=12515+15+1

Now, as you can see, the numerator is 12(sqrt(5)+1)12(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(5+1)(51)=(5)2+12=51=4

So, the fraction becomes

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

May 28, 2018

3sqrt5+335+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)1251×5+15+1

rArr (12sqrt5+12)/4125+124

rArr 3sqrt5+335+3