SOLUTION: I really need some help. the square root of 3x^4y^3 divided the sqaure root of 9xy^5 The only thing I thought to do was to put the problems vertical instead of horizontal and then

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Question 142299: I really need some help. the square root of 3x^4y^3 divided the sqaure root of 9xy^5 The only thing I thought to do was to put the problems vertical instead of horizontal and then get the square root of 9. Im not sure what the next step is.
Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(48535) About Me  (Show Source):
You can put this solution on YOUR website!
square root of 3x^4y^3 divided the sqaure root of 9xy^5
-----------
= sqrt[3x^4y^3/9xy^5]
Multiply numerator and denominator by xy so the denominator becomes
a perfect square:
= sqrt[3x^5y^4/9x^2y^6]
Simplify the denominator:
= sqrt[3x^5y^4] / 3xy^3
Simplify the numerator:
= [x^2y^2/3xy^3] sqrt(3x)
Reduce the fraction:
= [x/3y]*sqrt(3x)
===================
Cheers,
Stan H.

Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%283x%5E4y%5E3%29%2Fsqrt%289xy%5E5%29+

First, make a single square root:
sqrt%28%283x%5E4y%5E3%29%2F%289xy%5E5%29%29+

Next, reduce the fraction, but I recommend NOT reducing the 3 and the 9, since you prefer to have a perfect square of 9 in the denominator. Do it this way:
sqrt%28%283x%5E3%29%2F%289y%5E2%29%29+

The perfect square denominator is great. Take this square root:
%28sqrt%283x%5E3%29+%29%2F%283y%29+

Simplify the numerator by breaking it down into two square roots, placing all perfect squares in the first square root sign.
%28sqrt%28x%5E2%29%2Asqrt%283x%29+%29%2F%283y%29+
%28x%2Asqrt%283x%29+%29%2F%283y%29+

YOU HAVE GOT TO SEE MY WEBSITE FOR A SIMPLER EXPLANATION OF SQUARE ROOTS!! Click on my tutor name "rapaljer" anywhere in algebra.com. Take the second link on my homepage, which is "Basic, Intermediate, and College Algebra: One Step at a Time." Choose Basic Algebra, and look for Chapter 5. I think it will be great support for whatever textbook you are using!!

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