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

2 Answers
May 28, 2018

Answer:

#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)#

May 28, 2018

Answer:

#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#