What is #sqrt(80xy^2z)#?

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Answer 1 · 2017-06-14T20:11:58

#4|y|sqrt(5xz)#

To simplify this, we need to use two important properties of square roots:

1. #sqrt(a*b) = sqrta * sqrtb#

2. #sqrt(a^2) = |a|#

First, let's break #80# into prime factors.

#80 = color(red)2 * 40#
#40 = color(red)2 * 20#
#20 = color(red)2 * 10#
#10 = color(red)2 * color(red)5#

So #80 = color(red)2 * color(red)2 * color(red)2 * color(red)2 * color(red)5#

Using the properties above, we can see that:

#sqrt(80xy^2z)#

#= sqrt(2*2*2*2*5*x*y^2*z)#

We want to use the first rule to "extract" the perfect squares, and then the second rule to turn them into non-radical numbers.

#= sqrt(2*2) * sqrt(2*2) * sqrt(y^2) * sqrt(5*x*z)#

#2*2# is #2^2#, so this becomes:

#= |2| * |2| * |y| * sqrt(5xz)#

#=4|y|sqrt(5xz)#

Final Answer