hogie
Simplifying radical expressions.
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hogie
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\[\sqrt{4a^2}\]
hogie
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|dw:1366344988992:dw|
hogie
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What do i do with the a^2
hogie
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?
hogie
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dave0616
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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
hogie
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|dw:1366345329452:dw|
dave0616
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same principle is used. (4x^4)^1/2, 4*(1/2)=2= 2x^2
hogie
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dave0616
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correct
hogie
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dave0616
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i wanna help you through this one. what is x*x*x*x?
hogie
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|dw:1366346145214:dw|
hogie
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you can only pull two x's out right?
dave0616
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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
dave0616
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so you have x^2
hogie
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|dw:1366346262015:dw|
hogie
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That would be the answer? ^^^
dave0616
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nope, because you origonally had :\[\sqrt{x^{4}}\]
and you took the square root of that to get x^2
hogie
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So what would it be? Just x^2
hartnn
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\(\large \sqrt{9x^4}= 3x^2\)
is correct.
dave0616
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@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
hartnn
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|dw:1366347228041:dw|
dave0616
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ohhhh i mistook that as a four, im sorry, then yes 3x^2 would be correct