Question

SIMPLIFY
square root of 20^4

Answer 1

Do you know the rules for rational exponents?

\[\Large \sqrt{20^4}= (20^4)^{1/2}\]

Answer 2

or, alternatively....

\[\Large \sqrt{20^4}=\sqrt{(20^2)^2}\]

Answer 3

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

Answer 4

sorry wasn't sure if it made sense..

Answer 5

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. :)

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

\[\Large \sqrt{20x^4}=\sqrt{20}\cdot \sqrt{x^4}=\sqrt{4\cdot 5}\cdot \sqrt{(x^2)^2}\]

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.

Answer 10

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

Answer 11

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 \(\sqrt{20}\)?

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

wouldn't it just be x?

Answer 17

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.

Answer 18

so x^2?

Answer 19

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

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

Answer 20

RIGHT! You got it!

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

ohh okay! Got it!

Answer 26

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

Answer 27

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Answer 28

:)

Answer 29

thank you so much!!!

Answer 30

You're welcome, happy to help. :)