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Thread: Quadratic equation!

  1. #1
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    Quadratic equation!

    Find [m] such that the sum of solutions of the equation
    $\displaystyle
    x^2+(2+m-m^2)*x+m^2=0
    $
    is equal to 0, and the product is equal to 4.

    My solution:

    $\displaystyle
    x1+x2=-\frac{b}{a}
    $
    $\displaystyle
    x1*x2=\frac{c}{a}
    $
    $\displaystyle
    x1+x2=-(2+m-m^2)
    $
    $\displaystyle
    x1*x2=m^2
    $
    $\displaystyle
    -(2+m-m^2)=0
    $
    $\displaystyle
    m^2=4
    $
    $\displaystyle
    m=\sqrt{4}
    $
    $\displaystyle
    m=2
    $

    But $\displaystyle x^2+(2+2-2^2)*x+2^2=0 $ no real solution.
    How to solve this?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Alex2103 View Post
    Find [m] such that the sum of solutions of the equation
    $\displaystyle
    x^2+(2+m-m^2)*x+m^2=0
    $
    is equal to 0, and the product is equal to 4.

    My solution:

    $\displaystyle
    x1+x2=-\frac{b}{a}
    $
    $\displaystyle
    x1*x2=\frac{c}{a}
    $
    $\displaystyle
    x1+x2=-(2+m-m^2)
    $
    $\displaystyle
    x1*x2=m^2
    $
    $\displaystyle
    -(2+m-m^2)=0
    $
    $\displaystyle
    m^2=4
    $
    $\displaystyle
    m=\sqrt{4}
    $
    $\displaystyle
    m=2
    $

    But $\displaystyle x^2+(2+2-2^2)*x+2^2=0 $ no real solution.
    How to solve this?
    Nothing in the problem statement says that the solutions need to be real. Looks good to me.

    -Dan
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  3. #3
    MHF Contributor

    Joined
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    $\displaystyle (x- a)(x-b)= x^2- (a+ b)x+ ab$

    If a and b are the roots of $\displaystyle x^2+ (2+ m+ m^2)x+ m^2= 0$, then $\displaystyle a+ b= -2- m- m^2$ and $\displaystyle ab= m^2$.

    If you require that "the sum of the roots is equal to 0, and the product is equal to 4" then you must have [itex]-2- m- m^2= 0[/tex] and [tex]m^2= 4[tex]. $\displaystyle m^2+ m+ 2= 0$ has two roots and one of them will satisfy $\displaystyle m^2= 4$.
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