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Thread: Completing the Square problem solving

  1. #1
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    Completing the Square problem solving

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

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

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

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

    Whats up with that?
    The answers are identical.

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

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

    $\displaystyle b = \sqrt{b} \cdot \sqrt{b}$

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

    $\displaystyle b = \sqrt{b} \cdot \sqrt{b}$

    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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