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Math Help - Can someone look this over please?

  1. #1
    Junior Member SuperCalculus's Avatar
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    Can someone look this over please?

    So, I was given the task of proving the quadratic formula for homework... I think I've pretty much got it, but I'd like to be sure, so if someone can look over my working it'd be much appreciated.

    ax^2 + bx + c = 0

    Completing the square gives;

    a(x + \frac{b^2}{2a})^2 - \frac{b^2}{4a} + c = 0

    a(x + \frac{b^2}{2a})^2 = \frac{b^2}{4a} - c

    a(x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a}

    (x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a^2}

    (x + \frac{b^2}{2a})  = \frac{\sqrt {b^2 - 4ac}}{\sqrt{4}\sqrt{a^2}} = \frac{\sqrt {b^2 - 4ac}}{2a}

    x = - \frac{b}{2a} \pm \frac{\sqrt {b^2 - 4ac}}{2a} = \frac{-b\pm \sqrt {b^2 - 4ac}}{2a}
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by SuperCalculus View Post
    So, I was given the task of proving the quadratic formula for homework... I think I've pretty much got it, but I'd like to be sure, so if someone can look over my working it'd be much appreciated.

    ax^2 + bx + c = 0

    Completing the square gives;

    a(x + \frac{b^2}{2a})^2 - \frac{b^2}{4a} + c = 0

    a(x + \frac{b^2}{2a})^2 = \frac{b^2}{4a} - c

    a(x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a}

    (x + \frac{b^2}{2a})^2 = \frac{b^2 - 4ac}{4a^2}

    (x + \frac{b^2}{2a})  = \frac{\sqrt {b^2 - 4ac}}{\sqrt{4}\sqrt{a^2}} = \frac{\sqrt {b^2 - 4ac}}{2a}

    x = - \frac{b}{2a} \pm \frac{\sqrt {b^2 - 4ac}}{2a} = \frac{-b\pm \sqrt {b^2 - 4ac}}{2a}
    Pretty much but the \pm sign should also be in the second last step

    interestingly it's different to how I derive it still correct ^_^
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  3. #3
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Pretty much but the \pm sign should also be in the second last step

    interestingly it's different to how I derive it still correct ^_^
    Thank you ^.^. How do you derive it?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by SuperCalculus View Post
    Thank you ^.^. How do you derive it?
    I divide by a initially

    ax^2+bx+c = 0

    x^2 + \frac{b}{a}x + \frac{c}{a} = 0

    \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0

    <br />
\left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}

    Then as you did it
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