How do you solve #sqrt(48x^3y^2)/sqrt( 4xy^3)#?

2 Answers
Apr 20, 2017

Answer:

#sqrt(48xy^2x^2)/sqrt(4xy^2y)# final answer: #(2xsqrt(3))/sqrt(y)#

Explanation:

Therefore simplfiy the above equation:

#sqrt(12x^2)/sqrt(y)# or

#(2sqrt(3x^2))/sqrt(y)# or rearranging

#(2xsqrt(3))/sqrty#

Apr 20, 2017

Answer:

#(2*xsqrt(3))/(sqrt(y))#

Explanation:

Square roots are scary looking, but they're not that bad. You just need to look for pairs of numbers (or three numbers if it's a cube root).
I'll show you with an easy example:
#sqrt(9)#
#sqrt(3*3)#
#sqrt(3^2)#
#3#

When there are pairs of numbers being multiplied under the square root, it means there are squares under the square root.
Squares and square roots are inverses of each other, so they cancel each other out.

With our expression, our first job is to expand everything:
#(sqrt(48x^3y^2))/(sqrt(4xy^3))# becomes
#(sqrt(2*2*2*2*3*x*x*x*y*y))/(sqrt(2*2*x*y*y*y))#

Now we need to look for pairs. I'm going to group them to make them easier to see

#(sqrt(color(red)(2*2)*color(red)(2*2)*3*color(red)(x*x)*x*color(red)(y*y)))/(sqrt(color(red)(2*2)*x*color(red)(y*y)*y))#

Now let's rewrite these pairs as squared values:

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

All of the squares are moved outside of the square root:

#(2*2*x*ysqrt(3*x))/(2*ysqrt(x*y))#

We can simplify the numbers that are outside of the square root:

#(cancel(2)*2*x*cancel(y)sqrt(3*x))/(cancel(2)*cancel(y)sqrt(x*y))#
#(2*xsqrt(3*x))/(sqrt(x*y))#

And we can simplify the square roots too!

Let's expand the square roots:

#(2*xsqrt(3)sqrt(x))/(sqrt(x)sqrt(y))#

Simplify

#(2*xsqrt(3)cancel(sqrt(x)))/cancel(sqrt(x)sqrt(y))#
#(2*xsqrt(3))/(sqrt(y))#

We can move the #sqrty# out of the denominator, but if we leave the expression as it is we should be fine.