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?

2. Originally Posted by Alex2103
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

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

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 $\displaystyle m^2= 4$.