A nice quadratic ^^

**Moo** · November 7th 2008, 09:34 AM

Hello,

What's so special about this polynomial ? :

$P_m(x)=(m-3)x^2+(2m-1)x+(m+2)$ , $m \in \mathbb{R}$

The discriminant is $\Delta=(2m-1)^2-4(m-3)(m+2)=4m^2-4m+1-4(m^2-m-6)=\boxed{25}$

One can see that the discriminant is independent from m !

The solutions are :

$x_1=\frac{-(2m-1)-5}{2(m-3)}$ and $x_2=\frac{-(2m-1)+5}{2(m-3)}$

Hey, what else ??
Let's see $x_1$ :

$x_1=\frac{-2m+2-5}{2(m-3)}=\frac{-2(m-3)}{2(m-3)}=\boxed{-1}$ !!!

So $-1$ is always a root of this quadratic polynomial, whatever m is