How do you simplify #sqrt(108x^5y^6)#?

1 Answer
NJ
Apr 19, 2018

If you only want integer powers, then the following is simplified:

#=>6y^3sqrt(3x^5)#

If you are okay with fractional powers, then you could equivalently write:

#=>(6sqrt(3))x^(5/2)y^3#

Explanation:

We are given:

#sqrt(108x^5y^6)#

First, let's deal with the constant term #108#. We look for factors of #108# that are perfect squares, as this will allow us to pull it out of the radical.

Factors of #108 = {color(blue)1,2,3,color(blue)4,6,color(blue)9,12,18,27,color(blue)36,54,108}#.

The factors in #color(blue)"blue"# are perfect squares. To simplify the most, we want the largest one. So we choose #36#, which multiplied by #3# gives #108#.

#sqrt((36)(3) x^5y^6)#

#=> sqrt((6^2)(3)x^5y^6)#

#=> 6sqrt(3x^5y^6)#

We can also simplify the variable powers. Taking the square root is equivalent to raising to the #1/2# power. If you only want integer powers, then the following is simplified:

#6y^(6/2)sqrt(3x^5)#

#=>6y^3sqrt(3x^5)#

If you are okay with fractional powers, then you could equivalently write:

#=>(6sqrt(3))x^(5/2)y^3#

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