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Math Help - quadratic function

  1. #1
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    quadratic function

    How do you show that in the quadratic function: ax^2 + bx + c with 2 roots x1 and x2 that:

    1) x1 + x2 = -b/a

    and

    2) x1*x2 = c/a

    ?
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  2. #2
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    Quote Originally Posted by JQ2009 View Post
    How do you show that in the quadratic function: ax^2 + bx + c with 2 roots x1 and x2 that:

    1) x1 + x2 = -b/a

    and

    2) x1*x2 = c/a

    ?
    Use the fact that x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}. What happens when you add up the two possible answers?
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  3. #3
    Newbie I4talent's Avatar
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    Quote Originally Posted by JQ2009 View Post
    x_{1}+x_{2}= -\frac{b}{a}
    Let x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}

    and x_{2} = \frac{-b-\sqrt{b^2-4ac}}{2a}

    Then calculate x_{1}+x_{2}= \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)+\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)


    Alternatively:
    Spoiler:
    If x_{1} and x_{2} are the roots of the equation f(x) = ax^2+bx+c, then the equation must be of the form

    f(x) = k(x-x_{1})(x-x_{2}) for some constant k.

    Therefore, we have:
    k(x-x_{1})(k-x_{2}) \equiv ax^2+bx+c

    \implies k(x^2-(x_{1}+x_{2})x+x_{1}x_{2} \equiv ax^2+bx+c

    kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c<br />

    Equating the coefficients of x gives:
    -k(x_{1}+x_{2}) = b \implies x_{1}+x_{2} = -\frac{b}{k}

    But  k = a, so:

    \boxed{x_{1}+x_{2} = -\frac{b}{a}}


    Quote Originally Posted by JQ2009 View Post
    \left(x_{1}\right)\left(x_{2}\right) = \frac{c}{a}
    Let x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}

    and x_{2} = \frac{-b-\sqrt{b^2-4ac}}{2a}

    Then calculate \left(x_{1}\right)\left(x_{2}\right) = \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)

    Alternatively:
    Spoiler:
    If x_{1} and x_{2} are the roots of the equation f(x) = ax^2+bx+c, then the equation must be of the form

    f(x) = k(x-x_{1})(x-x_{2}) for some constant k.

    Therefore, we have:
    k(x-x_{1})(k-x_{2}) \equiv ax^2+bx+c

    \implies k(x^2-(x_{1}+x_{2})x+x_{1}x_{2} \equiv ax^2+bx+c

    kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c<br />

    Equating the constants gives:
    kx_{1}x_{2} = c \implies x_{1}x_{2} = \frac{c}{k}

    But  k = a, so:

    \boxed{x_{1}x_{2} = \frac{c}{a}}

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  4. #4
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    ax^2+bx+c = (X-X_1)(X-X_2)

    X^2-(X_1 + X_2)x + X_1X_2

    then you can plug in the values for a, b and c
    Last edited by sammy28; November 2nd 2009 at 08:51 AM. Reason: bad sign
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