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Thread: Completing the square with fraction as coefficient of x^2

  1. #1
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    Completing the square with fraction as coefficient of x^2

    Hi!

    Complete the square for $\displaystyle \frac12x^2-x+3$

    $\displaystyle \frac{-2}{2(1)}\,=\,1$

    $\displaystyle \frac12(x^2 - 2x + 1 - 1) + 3$

    $\displaystyle \frac12(x-1)^2 - 2$
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  2. #2
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    Re: Completing the square with fraction as coefficient of x^2

    You make life difficult on yourself if you skip steps. The first thing to do is to take out $\displaystyle \displaystyle \begin{align*} \frac{1}{2} \end{align*}$ as a factor, then complete the square on everything left over, then multiply that factor back through.
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    Re: Completing the square with fraction as coefficient of x^2

    $\displaystyle \frac{-2}{2(1)}\,=\,1$

    $\displaystyle \frac12(x^2 - 2x) + 3$

    $\displaystyle \frac12\left[(x^2 - 2x + 1 - 1) + 3\right]$

    $\displaystyle \frac12(x-1)^2 - 1 + 3$

    $\displaystyle \frac12(x-1)^2 + 2$
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  4. #4
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    Re: Completing the square with fraction as coefficient of x^2

    I'm not giving any more help until you learn to follow the instructions you've been given.
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  5. #5
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    Re: Completing the square with fraction as coefficient of x^2

    My solution is what I understand from the instructions given.

    Quote Originally Posted by Prove It
    The first thing to do is to take out $\displaystyle \frac12$ as a factor
    Factoring $\displaystyle \frac12$ from the $\displaystyle x^2$ and $\displaystyle x$ terms: $\displaystyle \frac12\left[(x^2 - 2x + 6)\right]$

    Quote Originally Posted by Prove It
    then complete the square on everything left over
    $\displaystyle \frac12(x^2 - 2x + 1 - 1 + 6)$

    $\displaystyle \frac12\left[(x-1)^2 + 5\right]$

    Quote Originally Posted by Prove It
    then multiply that factor back through.
    $\displaystyle \frac12(x-1)^2 + \frac52$
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  6. #6
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    Re: Completing the square with fraction as coefficient of x^2

    And this is correct. Well done.
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