How do you simplify #sqrt(81x^4y^9z^2)#?

3 Answers
Apr 7, 2015

Find the squares under the root and take them out (un-squared)

#sqrt(81x^4y^9z^2)=sqrt(9^2*(x^2)^2*(y^4)^2*y*z^2)=#

#9x^2y^4zsqrty#

(the odd #y# remains under the root)

Apr 7, 2015

To the extent possible factor the argument of the square root into squares:

#sqrt(81x^4y^9z^2)#

#= sqrt(9^2*(x^2)^2*(y^4)^2*y*z^2)#

#= 9x^2y^4zsqrt(y)#

Apr 7, 2015

9#x^2##y^4#z#sqrt(y)#

Except #y^9# all other terms are perfect squares, take square root of each of them and write outside the radical sign. Split #y^9# as y.#y^8#. Square root of #y^8# would be #y^4#. Write this outside the radical sign. The only term remaining inside the radical sign would be y. Hence the answer would be 9#x^2##y^4#z#sqrty#