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Math Help - Completing the Square using Fractions

  1. #1
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    Completing the Square using Fractions

     2x^2 -7x +3 = 0

    2x^2 -7x = -3

    \frac{\2x^2}{2} -\frac{\7}{2}x = -\frac{\3}{2}

    Take one half of - \frac{\7}{2}x and then square that answer which = \frac{\49}{16} and add to both sides

    x^2 -\frac{\7}{2} + \frac{\49}{16}= - \frac{\3}{2} + \frac{\49}{16}

    Combine like terms on the right to get

    \frac{\25}{16}

    and then subtract that from both sides to get

    x^2 -\frac{\7}{2} + \frac{\24}{16}=0

    \frac{\24}{16} can be reduced to \frac{\3}{2} so that

    x^2 -\frac{\7}{2} + \frac{\3}{2}=0

    Eliminate the fractions by multiplying it all by 2 and then instead of completing a square I've completed a large useless circle as I arrive back at the original equation.

    2x^2 -7x +3=0

    Now, I also used \frac{\49}{4} as the square of half of -\frac{\7}{2} but eventually it also ends up as
    \frac{\3}{2} on the left hand side of the equation.

    I did this problem with the quadratic formula in about 45 seconds but I have to learn it this way. If nothing else my LaTex skills have been markedly improved in this post.

    Any help is appreciated. Thanks.
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  2. #2
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    Re: Completing the Square using Fractions

    Quote Originally Posted by Ingersoll View Post
     2x^2 -7x +3 = 0

    2x^2 -7x = -3

    \frac{\2x^2}{2} -\frac{\7}{2}x = -\frac{\3}{2}

    Take one half of - \frac{\7}{2}x and then square that answer which = \frac{\49}{16} and add to both sides

    x^2 -\frac{7}{2}x +\frac{49}{16}= -\frac{3}{2}+\frac{49}{16}

    Combine like terms on the right to get

    \frac{\25}{16} ... ok

    and then subtract that from both sides to get ... no
    x^2 -\frac{7}{2}x+\frac{49}{16}= \frac{25}{16}

    \left(x - \frac{7}{4}\right)^2 = \frac{25}{16} this is why they call it "completing" the square

    "un" square both sides ...

    x - \frac{7}{4} = \pm \frac{5}{4}

    x = \frac{7 \pm 5}{4}

    btw ... one open tex at the beginning of a line ... one close /tex at the end of a line
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  3. #3
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    Re: Completing the Square using Fractions

    Your error occurs right about here:

    x^2 -\frac{\7}{2}x +\frac{\49}{16}= -\frac{\3}{2}+ \frac{\49}{16}

    Until this point, you had done everything right. Without doing anything else, the left-hand (only) can be factored into a product of squares. The right is a number, which we can find the square root of. First I'm going to rewrite this after simplifying the right-hand side.

    x^2 -\frac{\7}{2}x + \frac{\49}{16}= \frac{\25}{16}

    Now we can factor the left-hand side (AS IT IS RIGHT NOW).

    \left(x-\frac{7}{4}\right)^2=\frac{\25}{16}

    Now that it is in this form, we can take the square root of both sides. We are left with

    x-\frac{7}{4}=\pm \frac{5}{4} remember that every number has two square roots...

    x=\frac{7}{4}\pm \frac{5}{4}

    If we subtract, x=\frac{2}{4}=\frac{1}{2}. If we add, we have x=\frac{12}{4}=3. Thus x \in \{\frac{1}{2}, 3\}

    I go over the conceptual logic of completing the square in this .
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  4. #4
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    Re: Completing the Square using Fractions

    Thank you both so much. It's always right in front of my face. Thanks again.
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