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Mathematics 23 Online
OpenStudy (anonymous):

I GIVE MEDALS! Simplify by expressing with radicals. 6√x^2y^6

OpenStudy (luigi0210):

Is that whole thing under a radical or just the x?

OpenStudy (anonymous):

The whole thing is under the radical. @Luigi0210

OpenStudy (luigi0210):

Like?: \(\Large 6 \sqrt{x^2y^6}\)?

OpenStudy (anonymous):

\[6\sqrt{x^2y^6}=6\sqrt{x*x*y*y*y*y*y*y}=6xy^3\]

OpenStudy (luigi0210):

Have you learned your exponent rules yet? They help. @sujuexob2st Haha, I thought the same thing..

OpenStudy (anonymous):

Then when to multiply them, when to add, when to subtract, all that? @Luigi0210

OpenStudy (luigi0210):

Yes! Because \(\Large \sqrt[6]{x^2y^6}\) is the same as \(\Large (x^2y^6)^{1/6}\)

OpenStudy (anonymous):

Okay, yeah. But I don't understand how to simplify that again. @Luigi0210

OpenStudy (triciaal):

do like the ones we just did. keep x together and y together when you raise a power to a power multiply the exponents

OpenStudy (triciaal):

Because x2y6−−−−√6 is the same as (x2y6)1/6 x^2^1/6*y^6^1/6 x^2/6*y^6/6 you can finish

OpenStudy (luigi0210):

Well think of it this way, you know how \(\sqrt{64}\) is 8? That's because \(\sqrt{64}\) is the same as \((64)^{1/2}\). I think of the index as a way of telling you how many of something you need in order to take it out.. \(\sqrt{64}\) simplifies into \(\sqrt{8*8}\).. because of that, since there are two "8's" you can take it out to get just 8.. sorry terrible explanation >_<

OpenStudy (anonymous):

Okay, I think I followed that.

OpenStudy (luigi0210):

In yours for example.. the number required would be 6.. since there are only 2 x's, you can't take it out.. but there are 6 y's.. so you can take that out as \(\Large y\sqrt[6]{x^2}\) ..

OpenStudy (luigi0210):

This is why Im not a teacher :P

OpenStudy (anonymous):

Haha so wait, are you saying that's the answer?

OpenStudy (triciaal):

@Luigi0210 I think your error is that you are relating everything to square root only when the power is 1/2 and thus we use 2s when you have the power 1/6 you are finding what number multiplied by itself 6 times will give that value under the radical what you expressed is correct I used your post now to simplify x^2^1/6 = x^2/6 = x^1/3 y^6^1/6 = y ^6/6 = y^1 = y solution x^1/3*y

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