# Thread: Real roots

1. ## Real roots

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,

2. Hi

Originally Posted by sancellX
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,
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 ?

3. 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.

4. Originally Posted by bobak
evaluate the discriminate $\displaystyle b^2 - 4ac$

$\displaystyle 4^2(a-b)^2 - 4(a-b+c)(a-b-c)$
$\displaystyle 4^2(a-b)^2 - 4(a^2 -2ab +b^2 -c^2)$
$\displaystyle 4^2(a-b)^2 - 4(a-b)^2 +{\color{red}4}c^2$
$\displaystyle 12(a-b)^2+{\color{red}4}c^2$

The sum of two square is clearly positive.

Bobak