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Math Help - Completing the square with an uneven b

  1. #1
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    Completing the square with an uneven b

    I'm working through some exercises to brush up my pre-calculus algebra and I'm asked to solve several exercises by completing the square (n.b. using the abc-formula isn't allowed for these exercises).

    x^2 + 7x - 1 = 0

    This kind of puzzles me, again: I can't use the abc-formula. The uneven b (7x) doesn't allow for a square, unless I add an x, which would get me something like: x^2 + 8x + 16 = x + 17, which seems an unfruitful endeavor...

    Any tips?
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  2. #2
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    Re: Completing the square with an uneven b

    Quote Originally Posted by Lepzed View Post
    I'm working through some exercises to brush up my pre-calculus algebra and I'm asked to solve several exercises by completing the square (n.b. using the abc-formula isn't allowed for these exercises).

    x^2 + 7x - 1 = 0

    This kind of puzzles me, again: I can't use the abc-formula. The uneven b (7x) doesn't allow for a square, unless I add an x, which would get me something like: x^2 + 8x + 16 = x + 17, which seems an unfruitful endeavor...

    Any tips?
    The method is to add (b/2)^2 to both sides of the equation (when a = 1, as you have here).

    Then you'll have
    (x + b/2)^2 = ...

    So here, we add (7/2)^2 = 49/4 to both sides.

    That b is (or would be) even is just a luxury
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  3. #3
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    Re: Completing the square with an uneven b

    Quote Originally Posted by Lepzed View Post
    I'm working through some exercises to brush up my pre-calculus algebra and I'm asked to solve several exercises by completing the square (n.b. using the abc-formula isn't allowed for these exercises).

    x^2 + 7x - 1 = 0

    This kind of puzzles me, again: I can't use the abc-formula. The uneven b (7x) doesn't allow for a square, unless I add an x, which would get me something like: x^2 + 8x + 16 = x + 17, which seems an unfruitful endeavor...

    Any tips?
    \displaystyle \begin{align*} x^2 + 7x - 1 &= 0 \\ x^2 + 7x + \left(\frac{7}{2}\right)^2 - \left(\frac{7}{2}\right)^2 - 1 &= 0 \\ \left(x + \frac{7}{2}\right)^2 - \frac{49}{4} - \frac{4}{4} &= 0 \\ \left(x + \frac{7}{2}\right)^2 - \frac{53}{4} &= 0 \\ \left(x + \frac{7}{2}\right)^2 &= \frac{53}{4} \\ x + \frac{7}{2} &= \pm \frac{\sqrt{53}}{2} \\ x &= -\frac{7}{2} \pm \frac{\sqrt{53}}{2} \\ x &= \frac{-7 \pm \sqrt{53}}{2} \end{align*}
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    Re: Completing the square with an uneven b

    Quote Originally Posted by Lepzed View Post
    I'm working through some exercises to brush up my pre-calculus algebra and I'm asked to solve several exercises by completing the square (n.b. using the abc-formula isn't allowed for these exercises).

    x^2 + 7x - 1 = 0

    This kind of puzzles me, again: I can't use the abc-formula. The uneven b (7x) doesn't allow for a square, unless I add an x, which would get me something like: x^2 + 8x + 16 = x + 17, which seems an unfruitful endeavor...

    Any tips?
    Have you never taken an arithmetic course in which you learned to deal with fractions?
    7 divided by 2 is 7/2. It's square is 49/4. I would think that a person taking algebra would know that.
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  5. #5
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    Re: Completing the square with an uneven b

    I just didn't think of it.
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