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Thread: 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 $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. What happens when you add up the two possible answers?
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  3. #3
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    Quote Originally Posted by JQ2009 View Post
    $\displaystyle x_{1}+x_{2}= -\frac{b}{a}$
    Let $\displaystyle x_{1} = \frac{-b+\sqrt{b^2-4ac}}{2a}$

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

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


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

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

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

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

    $\displaystyle kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c
    $

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

    But $\displaystyle k = a$, so:

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


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

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

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

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

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

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

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

    $\displaystyle kx^2-(x_{1}+x_{2})kx+x_{1}x_{2} \equiv ax^2+bx+c
    $

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

    But $\displaystyle k = a$, so:

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

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

    $\displaystyle X^2-(X_1 + X_2)x + X_1X_2$

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