# Real roots

• Jun 19th 2008, 02:28 PM
sancellX
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,
• Jun 19th 2008, 02:35 PM
Moo
Hi :)

Quote:

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 ? :)
• Jun 19th 2008, 02:35 PM
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 +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.
• Jun 19th 2008, 02:36 PM
Moo
Quote:

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

(Tongueout)