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

2 Answers
Sep 3, 2017

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

Explanation:

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)

Sep 3, 2017

I different interpretation of the question to that by Tazwar Sikder

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

Explanation:

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")