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Math Help - Solving this simple function

  1. #1
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    Solving this simple function

    X^2-5x-3=0?

    I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!
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  2. #2
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    Quote Originally Posted by fattydq View Post
    X^2-5x-3=0?

    I honestly have no idea how to do this, perhaps x^2-5x=3, but then I'm still stuck, help!
    x^2-5x-3=0
    It doesn't factorise easily, do you know the quadratic formula?
    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} So in this case a=1 b=-5 and c=-3
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  3. #3
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    leave the equation equal to 0 to solve.

    since it can't be factorized you can solve either by

    \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}

    where a, b, c are the coefficients of the terms, or you can solve using 'completing the square'

    (x - \frac{5}{2})^2 - \frac{25}{4} - 3

    (x - \frac{5}{2})^2 = \frac{37}{4}

    x = \frac{5}{2} \pm \sqrt{\frac{37}{4}}
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  4. #4
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    Quote Originally Posted by JeWiSh View Post
    x^2-5x-3=0
    It doesn't factorise easily, do you know the quadratic formula?
    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} So in this case a=1 b=-5 and c=-3
    Oh yeah, I did already know that but for some reason my mind just blanked when I saw it. Thanks a bunch!
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