How do you simplify #sqrt(9/x^8) - sqrt(25/x^6)#?

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Answer 1 · 2016-01-28T15:38:44

Since each term under the square roots are perfect squares, we can get rid of the square roots completely.

-The square root of 9 is 3
-The square root of #x^8# is #x^4#
-The square root of 25 is 5
-The square root of #x^6# is #x^3#

So, #sqrt(9/x^8)# - #sqrt(25/x^6)# = #3/x^4# - #5/x^3#

Place on the same denominator by using exponent rules

#x^4# = #x^3 xx x^1#

As a result, we must multiply the numerator by #x^1# as well.

#(3 - 5x^1)/x^4#

= #(3 - 5x)/x^4#

Hopefully this helps.