SIMPLIFY
square root of 20^4

- anonymous

SIMPLIFY
square root of 20^4

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- DebbieG

Do you know the rules for rational exponents?
\[\Large \sqrt{20^4}= (20^4)^{1/2}\]

- DebbieG

or, alternatively....
\[\Large \sqrt{20^4}=\sqrt{(20^2)^2}\]

- DebbieG

lol while I was responding to your comment, you deleted it.... ?

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## More answers

- anonymous

sorry wasn't sure if it made sense..

- DebbieG

Well, what I was going to say is that that sounds like a convuluted approach :) but I presume that "prime number and perfect square" refers to writing 20 = 5*4, so I guess that means:
\[\Large \sqrt{20^4}=\sqrt{20\cdot 20\cdot 20\cdot 20}=\sqrt{20}\cdot\sqrt{20}\cdot\sqrt{20}\cdot\sqrt{20}\]
and then you can write each 20 as 5*4, so you can pull each 4 out from under the radical as the square root, 2, and then you have four factors of \(\Large \sqrt{5}\)....
Like I said, really convoluted. :)

- anonymous

oh my goodness.. i didn't add my x in there

- anonymous

it's supposed to be; the square root of 20x^4

- DebbieG

Oh, well that's very different indeed. lol.

- DebbieG

\[\Large \sqrt{20x^4}=\sqrt{20}\cdot \sqrt{x^4}=\sqrt{4\cdot 5}\cdot \sqrt{(x^2)^2}\]
Then simplify each part. you can pull out the 4, but not the 5. And you can completely simplify the radical involving the power of x.

- anonymous

what do you mean pull out the 4 but not 5?

- DebbieG

OK, to simplify a square root expression, you have to remember that you can "pull out" any perfect squares under the square root sign. Anything under there that ISNT a square doesn't get to come out and play, it has to stat where it is. :) I'll so a DIFFERENT one for you:
\[\Large \sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot \sqrt{2}=\sqrt{5^2}\cdot \sqrt{2}\]
Now, the 5^2 gets to come out, because the square root of a perfect square is just what you started with, right? The square root "undoes" the squaring.
But the 2 has to stay where it is, can't simplify because it isn't a perfect square... \(\sqrt{2}\) is just \(\sqrt{2}\), there is nothing more that you can do with it.
so:
\[\Large \sqrt{5^2}\cdot \sqrt{2}=5 \sqrt{2}\]
Now, can you do the same thing for \(\sqrt{20}\)?

- anonymous

i still don't fully understand how to do it for 20 bc of 4 and 5 :/

- DebbieG

Just like I did it for 50: factor to see what you have, under the sq root. Any perfect square comes out, because it simplifies (you take the sq. root - like I did above for 25)
\[\Large \sqrt{20}=\sqrt{4\cdot 5}=\sqrt{4}\cdot \sqrt{5}=\sqrt{2^2}\cdot \sqrt{5}\]
So, the \(\Large \sqrt{2^2}\) will simplify, since it's a perfect square, the 2 comes out. But the \(\Large \sqrt{5}\) stays under the radical, since 5 is not a perfect square.

- anonymous

\[2\sqrt{5x ^{4}}\]

- DebbieG

OK, you took care of the 20 part of it. Now what about the \(\Large x^4\)? That is a perfect square, since: \(\Large x^4=(x^2)^2\).

- anonymous

wouldn't it just be x?

- DebbieG

Not quite.... think about what is SQUARED under the SQUARE ROOT. Whatever it is that is squared under the root, that's what comes out.

- anonymous

so x^2?

- DebbieG

\[\Large z^6=(z^3)^2~~so~~\sqrt{ z^6}=z^3\]
\[\Large t^{20}=(t^{10})^2~~so~~\sqrt{ t^{20}}=t^{10}\]

- DebbieG

RIGHT! You got it!

- DebbieG

So now put it all together... what's the complete simplified form?

- anonymous

\[2\sqrt{5x ^{2}}\]

- DebbieG

Ack.... now wait a sec.... that \(x^2\) COMES OUT, remember?

- DebbieG

You "take the square root" so it comes OUT of the radical... out in front... just like the square root of 4 came out, so does the square root of x^4

- anonymous

ohh okay! Got it!

- anonymous

\[2x ^{2}\sqrt{5}\]

- DebbieG

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- DebbieG

:)

- anonymous

thank you so much!!!

- DebbieG

You're welcome, happy to help. :)

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