Answered

Can someone PLEASE explain how to simplify square roots with variables and eexponents in them?? I'd also be thankful if you explained step-by-step how to

solve (16x^4)^3/2 and how to simplify 3y^4/3 times 3yx^1/2? I have to do a retake on my math test and I REALLY need help!

Verified Answer :

x^a/b is  \(\sqrt[b]{x^a}\) . The way I memorise that is x^1/3 is the cubic root of x. Do you get it? In that case, x is raised to a power of 1 and the cubic root is practically has a power of 3.
In your example, 

\(\sqrt[ \frac{3}{2} ]{16 x^4}\) is practically square rooting each term then cubing them individually. Remember when square-rooting any index you halve it. I'll elaborate:

\(\sqrt{x^4}\) = \(x^{2}\)
\(\sqrt{16}\) = 4
Then cube each,
\(4^3\) = 64 
and \(( x^{2} )^3\) = \(x^{6}\)

As for the 2nd part: you must use the rules of indices.
\(x^{a} * x^{b} = x^{a+b}\)
So breaking the question up:

3 * 3 = 9
\(x^{ \frac{1}{2} }\) stays as is since the 2nd term does not contain x
now: 
\(y^{ \frac{4}{3} } * y^{1} = y^{ \frac{4}{3} + 1 } = y^{ \frac{4}{3} + \frac{3}{3} } = y^{ \frac{7}{3} }\)
This makes your final answer look like this:
\(9 x^{ \frac{1}{2} } y^{ \frac{7}{3} }\)

I hope that helped and good luck in your test!

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