Sqrt 128x^9y^16/[16x^2y]? #Simplify

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Answer 1 · 2017-09-03T09:25:48

#frac(sqrt(2 x^(5)) y^(7))(8)#

We have: #frac(sqrt(128 x^(9) y^(16)))(16 x^(2) y)#

#= frac(sqrt(128) cdot sqrt(x^(9)) cdot sqrt(y^(16)))(16 x^(2) y)#

#= frac(8 sqrt(2) cdot sqrt(x^(9)) cdot y^(8))(16 x^(2) y)#

#= frac(8 sqrt(2) y^(8) cdot sqrt(x^(9)))(16 x^(2) y)#

#= frac(sqrt(2) y^(7))(8) cdot (x^(frac(9)(2) - 2))#

#= frac(sqrt(2) y^(7))(8) cdot (x^(frac(5)(2)))#

#= frac(sqrt(2) y^(7))(8) cdot sqrt(x^(5))#

#= frac(sqrt(2 x^(5)) y^(7))(8)#

Answer 2 · 2017-09-03T09:49:07

I different interpretation of the question to that by Tazwar Sikder

#2x^3y^7sqrt(2xy) color(white)(ddd)# Answer checked and confirmed

Assumption: the question is meant to read:

#sqrt((128x^9y^16)/(16x^2y)#

If you are ever not sure about the roots of a large number do a rough sketch of a prime factor tree.
Tony B

Write as:

#(sqrt(128x^9y^16))/(sqrt(16x^2y))#

This is the same as

#(sqrt((2^2)^3xx2xx (x^2)^4xx x xx(y^2)^8))/(sqrt(4^2xx x^2xxy)#
#color(white)("d")#

#(2^3x^4y^8sqrt(2x))/( 4xsqrt(y)) #

Cancel out the #4x# in the denominator

#(2x^3y^8sqrt(2x))/( sqrt(y)) #

Multiply by 1 but in the form of #1=sqrt(y)/sqrt(y)#

#(2x^3y^8sqrt(2xy))/y #

Cancel out the #y# in the denominator

#2x^3y^7sqrt(2xy) #
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check by substituting arbitrary values for #x and y#

I choose #x=2; y=2#

#sqrt((128x^9y^16)/(16x^2y)=5792.61875.....#

#2x^3y^7sqrt(2xy) =5792.61875.... color(red)(larr" It works")#