# A nice quadratic ^^

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• Nov 7th 2008, 09:34 AM
Moo
A nice quadratic ^^
Hello,

What's so special about this polynomial ? :

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

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

One can see that the discriminant is independent from m !

The solutions are :

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

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

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

So $\displaystyle -1$ is always a root of this quadratic polynomial, whatever m is (Tongueout)

This, however, can be explained : $\displaystyle (2m-1)-\bigg[(m-3)+(m+2)\bigg]=0$, so -1 is a root.

Isn't it nice ? (Surprised)
• Nov 7th 2008, 10:19 AM
ThePerfectHacker
Quote:

Originally Posted by Moo
Hey, what else ??

Hey

Quote:

Let's see $\displaystyle x_1$ :

$\displaystyle x_1=\frac{-2m+2-5}{2(m-3)}=\frac{-2(m-3)}{2(m-3)}=\boxed{-1}$ !!!
Why? (Worried)
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By the way, $\displaystyle m\not = 3$ (Wink)