How do you simplify #(6-sqrt20) / 2#?

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Answer 1 · 2016-02-24T16:04:16

#3-sqrt5#

First, recognize that we can simplify #sqrt20#, since #20=4xx5#.

We can split up a square root through the rule that

#sqrt(axxb)=sqrtasqrtb#

So,

#sqrt20=sqrt(4xx5)=sqrt4sqrt5=2sqrt5#

Thus, the expression equals

#(6-2sqrt5)/2#

We can split up the fraction:

#6/2-(2sqrt5)/2#

Which equals

#3-sqrt5#

Answer 2 · 2016-02-26T11:30:32

A very slight variation in presentation. Also written with a lot of detail about each step.

#" "3-sqrt(5)#

Looking for common factors. 6 and 20 are even so have a factor of 2. As the denominator is 2 as well we have a first step in simplification

Write as: #" " 6/2 -sqrt(20)/2#

#" "(2xx3)/2-(sqrt(2xx10))/2#

But #2xx5 = 10# so we now have

#" "((2xx3)/2)-((sqrt(2^2xx5))/2)#

#" "(2/2 xx 3)-((2sqrt(5))/2)#

#" "(1 xx 3) -(2/2xxsqrt(5))#

#" "(1 xx 3) -(1xxsqrt(5))#

#" "3-sqrt(5)#