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Math Help - Completing the Square problem solving

  1. #1
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    Completing the Square problem solving

    3p^2-12p+2=0
    p^2-4p+\frac{2}{3}=0
    Completing the square
    (\frac{4}{2})^2=4 and \frac{2}{3}+x=4 so add \frac{10}{3} to each side.
    p^2-4p+4=\frac{10}{3} which is (p-2)^2=\frac{10}{3}
    I end up with  p=2 \pm \sqrt{\frac{10}{3}}

    The book says \frac{6 \pm \sqrt{30}}{3}

    Whats up with that?
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  2. #2
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    Quote Originally Posted by bkbowser View Post
    3p^2-12p+2=0
    p^2-4p+\frac{2}{3}=0
    Completing the square
    (\frac{4}{2})^2=4 and \frac{2}{3}+x=4 so add \frac{10}{3} to each side.
    p^2-4p+4=\frac{10}{3} which is (p-2)^2=\frac{10}{3}
    I end up with  p=2 \pm \sqrt{\frac{10}{3}}

    The book says \frac{6 \pm \sqrt{30}}{3}

    Whats up with that?
    The answers are identical.

    \frac{6 \pm \sqrt{30}}{3} = 2 \pm \frac{\sqrt{30}}{3} = 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}} = 2 \pm \sqrt{\frac{10}{3}}
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    The answers are identical.

    \frac{6 \pm \sqrt{30}}{3} = 2 \pm \frac{\sqrt{30}}{3} = 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}} = 2 \pm \sqrt{\frac{10}{3}}
    OK why does the 2 \pm \frac{\sqrt{3}\sqrt{10}}{\sqrt{3}\sqrt{3}} section work like that? I get the nominator portion but not the denominator.
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  4. #4
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    \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

    b = \sqrt{b} \cdot \sqrt{b}

    Can you see why now?
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  5. #5
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    Quote Originally Posted by Defunkt View Post
    \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

    b = \sqrt{b} \cdot \sqrt{b}

    Can you see why now?
    Ya, since three can be written as a product of the square root of 3 and the square root of three.

    Thats pretty cool by the way, never really thought about doing that.
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