What is the simplification of #sqrt((18x^5y^4)/(49xz^3))#?

1 Answer
Jul 31, 2016

Answer:

#(3x^2y^2sqrt(2z))/(7z^2)#

Explanation:

#color(blue)("General comment")#

Any values that are squares can be taken outside the square root. So for example if you had #sqrt(2^2xx3)# you could write this as
#2sqrt(3)#. Or if you had #sqrt(2^4xx3)# you could write #2^2sqrt(3)#

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#color(blue)("Answering the question")#

Write as #sqrt(2xx3^2xx x xx x^4xx y^4)/(sqrt(7^2 xz xx z^2)# giving:

#(3x^2y^2sqrt(2x))/(7zsqrt(xz)) #

Write as:#" "(3x^2y^2sqrt2cancel(sqrtx))/(7zsqrtz cancel(sqrtx)) #

Multiply by 1 but in the form of #1=sqrtz/sqrtz# giving

#(3x^2y^2)/(7z)xxsqrt(2)/sqrt(z) xx sqrtz/sqrtz #

#(3x^2y^2sqrt(2z))/(7z^2)#