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Simplifying radical expressions.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\sqrt{4a^2}\]
|dw:1366344988992:dw|
What do i do with the a^2

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Other answers:

?
|dw:1366345074437:dw|
a is squared, and the opposite of a square root is to the second power. because this expression can also be written as (4a^2)^1/2. so if you take 2 and multiply it by a half, you get 1. so the answer is 2a
|dw:1366345329452:dw|
same principle is used. (4x^4)^1/2, 4*(1/2)=2= 2x^2
|dw:1366345558930:dw|
correct
|dw:1366345964027:dw|
i wanna help you through this one. what is x*x*x*x?
|dw:1366346145214:dw|
you can only pull two x's out right?
that is correct. so now you have sqrt(x^4). well in a way yes, but lets go ahead and use the formula we had.: (x^4)^1/2 4*(1/2)=2
so you have x^2
|dw:1366346262015:dw|
That would be the answer? ^^^
nope, because you origonally had :\[\sqrt{x^{4}}\] and you took the square root of that to get x^2
So what would it be? Just x^2
\(\large \sqrt{9x^4}= 3x^2\) is correct.
@hartnn and @hogie i guess i am confused as to where the sqrt(9x^4) came from, because i dont see that any where, thought 3x^2 would be correct for that
|dw:1366347228041:dw|
ohhhh i mistook that as a four, im sorry, then yes 3x^2 would be correct

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