Hi,
Prove that the roots of the following equation are real:
$\displaystyle (a-b+c)x^2 + 4(a-b)x + (a-b-c) = 0 $
Thanks,
Hi
This equation has real solutions iff $\displaystyle \Delta \ge 0$, where $\displaystyle \Delta$ is the discriminant.
$\displaystyle mx^2+nx+p=0 \implies \Delta=n^2-4mp$
Therefore :
$\displaystyle \Delta=[4(a-b)]^2-4(a-b+c)(a-b-c)$
$\displaystyle \Delta=16(a-b)^2-4{\color{red}[(a-b)+c][(a-b)-c]}$
The red part is something like $\displaystyle (x+y)(x-y)=x^2-y^2$, with x=(a-b) and y=c.
So $\displaystyle \Delta=16(a-b)^2-4[(a-b)^2-c^2]=16(a-b)^2-4(a-b)^2+4c^2=12(a-b)^2+4c^2$
Knowing that a squared number is always positive, what can you conclude ?
evaluate the discriminate $\displaystyle b^2 - 4ac$
$\displaystyle 4^2(a-b)^2 - 4(a-b+c)(a-b-c)$
$\displaystyle 4^2(a-b)^2 - 4(a^2 -2ab +b^2 -c^2)$
$\displaystyle 4^2(a-b)^2 - 4(a-b)^2 +4c^2$
$\displaystyle 12(a-b)^2+4c^2$
The sum of two square is clearly positive.
Bobak
Edit: I got beaten by a matter of seconds.