Square Root of Expressions containing Variables

Jackson332006

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Oct 19, 2006
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I am not sure how to simplify this. I try to use my book but I just don't understand.

(sqrt)100x^2y^16z^8
 
sqrt[100x^2y^16z^8]

we will do it in steps
sqrt[100times q]=10sqrt q then
sqrt[100x^2y^16 z^8=10 sqrt[x^2y^16z^8]

but sqrt x^2=x or we have
10x sqrt[y^16 z^8] but
y^16 = y^8 y^8 because you add exponent when you multiply. or we have
10 x y^8 sqrt[z^8]
10 x y^8 z^4 answer

a simple check on sqrt[x^6] = x^3 check to see if x^6=x^3 x^3 because you add exponents
let x=3 then 3^6=729 sqrt 729=27 which is 3^3

Arthur
 
Square roots can be broken up (ONLY if there is a single term under the root, which there is in this case).

\(\displaystyle \sqrt{100x^2y^{16}z^8} = \sqrt{10^2} \sqrt{x^2} \sqrt{y^{16}} \sqrt{z^8}\)

Then, you can break higher powers into squares:

\(\displaystyle \sqrt{10^2} \sqrt{x^2} \sqrt{y^{16}} \sqrt{z^8} = \sqrt{10^2} \sqrt{x^2} \sqrt{(y^{8})^2} \sqrt{(z^4)^2}\)

Now you can cancel call the squares with the square roots:

\(\displaystyle \sqrt{10^2} = 10\\ \sqrt{x^2} = |x| \text{ (if you are assuming x is positive you may omit the absolute value)} \\ \sqrt{(y^{8})^2} = y^8 \\ \sqrt{(z^4)^2} = z^4\)

So,

\(\displaystyle \sqrt{10^2} \sqrt{x^2} \sqrt{(y^{8})^2} \sqrt{(z^4)^2} = 10*|x|y^8z^4\)
 
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