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Math Help - Quadratic equation!

  1. #1
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    Quadratic equation!

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

    My solution:

    <br />
x1+x2=-\frac{b}{a} <br />
        <br />
x1*x2=\frac{c}{a}     <br />
    <br />
x1+x2=-(2+m-m^2)  <br />
     <br />
x1*x2=m^2<br />
    <br />
-(2+m-m^2)=0<br />
    <br />
m^2=4<br />
    <br />
m=\sqrt{4}<br />
    <br />
m=2<br />

    But 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
    <br />
x^2+(2+m-m^2)*x+m^2=0<br />
    is equal to 0, and the product is equal to 4.

    My solution:

    <br />
x1+x2=-\frac{b}{a} <br />
        <br />
x1*x2=\frac{c}{a}     <br />
    <br />
x1+x2=-(2+m-m^2)  <br />
     <br />
x1*x2=m^2<br />
    <br />
-(2+m-m^2)=0<br />
    <br />
m^2=4<br />
    <br />
m=\sqrt{4}<br />
    <br />
m=2<br />

    But 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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    (x- a)(x-b)= x^2- (a+ b)x+ ab

    If a and b are the roots of x^2+ (2+ m+ m^2)x+ m^2= 0, then a+ b= -2- m- m^2 and 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]. m^2+ m+ 2= 0 has two roots and one of them will satisfy m^2= 4.
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