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Math Help - help simplify please

  1. #1
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    help simplify please

    How would I simplify (144a^8b^2)^1/2

    I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help
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  2. #2
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    Quote Originally Posted by Ronis
    How would I simplify (144a^8b^2)^1/2

    I didn't know how to make the 1 over 2 smaller so I tried my bestbut its suppose to be in the top right, thanks if you can help
    It means square root,
    \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b

    -----
    Note: (this is not for you) I know that \sqrt{b^2}\not = b, so do not bother me with that. Because in schools terms are positive.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    It means square root,
    \sqrt{144a^8b^2}=\sqrt{144}\cdot \sqrt{a^8}\cdot \sqrt{b^2}=12a^4b

    -----
    Note: (this is not for you) I know that \sqrt{b^2}\not = b, so do not bother me with that. Because in schools terms are positive.
    {s con/con} (lol) thanks =]
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    Note: (this is not for you) I know that \sqrt{b^2}\not = b, so do not bother me with that. Because in schools terms are positive.
    Did you think one of us would correct you? Tsk, tsk.

    BTW. to make an exponent with more than one number all you need to do is wrap it in { }

    P.S. I would've corrected you hacker
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  5. #5
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    Hello, Ronis!

    You have to enclose the entire exponent in braces
    . like this: (144a^8b^2)^{1/2}

    You can even write: (144^8b^2)^{\frac{1}{2}}


    How would I simplify (144a^8b^2)^{\frac{1}{2}}

    Each factor is raised to the power: . (144)^{\frac{1}{2}}(a^8)^\frac{1}{2}}(b^2)^{\frac{  1}{2}}

    Since 144 = 12^2, we have: . (12^2)^{\frac{1}{2}}(a^8)^{\frac{1}{2}}(b^2)^{ \frac{1}{2}}

    Use that third property of exponents: . 12^{(2\cdot\frac{1}{2})} \cdot a^{(8\cdot\frac{1}{2})}\cdot b^{(2\cdot\frac{1}{2})}

    . . And we get: . 12^1\cdot a^4\cdot b^1\;=\;12a^4b

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  6. #6
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    Just for reference...

    \sqrt[2]{x}=x^{\frac{1}{2}
    also,
    \sqrt[3]{x}=x^{\frac{1}{3}
    you can remember the rule,
    \sqrt[a]{x}=x^{\frac{1}{a}

    also the rules of powers:

    x^a\times x^b=x^{(a+b)}

    (x^a)^b=x^{(a\times b)}

    x^{-a}=\frac{1}{x^a}
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