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(75887)   (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(4671)   (Show Source): You can put this solution on YOUR website!


First, make a single square root:


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:


The perfect square denominator is great. Take this square root:


Simplify the numerator by breaking it down into two square roots, placing all perfect squares in the first square root sign.



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R^2

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