Quadratic equation!

**Alex2103** · March 12th 2011, 11:26 AM

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

My solution:

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

But $x^2+(2+2-2^2)*x+2^2=0$ no real solution.
How to solve this?

**topsquark** · March 12th 2011, 01:01 PM

Originally Posted by Alex2103

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

My solution:

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

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

**HallsofIvy** · March 12th 2011, 02:40 PM

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

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 $m^2= 4$ .

Thread: Quadratic equation!

LinkBack

Thread Tools

Display

Quadratic equation!

Similar Math Help Forum Discussions

some quadratic equation

Quadratic Equation

Need Quadratic Equation

Is there a cubic equation like the quadratic equation?

Quadratic Equation

Search Tags