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how to simplify a square root?
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the equation i am doing is \[2\sqrt{3}-4\]
|dw:1340300701868:dw|
Depends... \[\sqrt{4}=\sqrt{2*2}=2\] \[\sqrt{12}=\sqrt{2*2*3}=2\sqrt{3}\] \[\sqrt{-1}=i\] \[\sqrt{-8}=\sqrt{2*2*2*-1}=2i\sqrt{2}\]
my answer should look like this (__√__)/__
For this you're using the Pythagorean Theorem a^2+b^2=c^2 where c = hypotenuse
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yes, but i don't know how I find the answer I need.
\[2^2+x^2=(2\sqrt{3})^2\]
Oh! I forgot to square the 2√3!
solving for x here, and remember that squaring a square root cancels it out because: \[\sqrt{x}=x^{1/2}\]
\[(x^{1/2})^2=x^{2/2}=x^1=x\]
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Got it? :D
yeah, Thanks.
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