# Real roots

• Jun 25th 2008, 02:22 PM
Recklessid
Real roots
Hi, I worked out a solution for this problem but I need someone to check it:

Prove that the roots of $\displaystyle (x-a)(x-b) = h^2$ are always real:

$\displaystyle x^2 -bx-ax+ab-h^2 =0$
$\displaystyle A= 1, B= -(a+b), C= ab-h^2$ where $\displaystyle B^2 - 4AC$ = positive or zero for the roots to be real.

$\displaystyle B^2-4AC = [-(a+b)]^2 -4(1)(ab-h^2)$

$\displaystyle a^2+2ab+b^2-4ab+4h^2 = a^2-2ab+b^2 +4h^2$

$\displaystyle = (a-b)^2 + 4h^2$

$\displaystyle (a-b)^2+4h^2$ consists of two perfect squares, which can never be negative. Specifically, the roots are rational because $\displaystyle B^2- 4AC$ is a perfect square.
• Jun 25th 2008, 02:30 PM
Jhevon
Quote:

Originally Posted by Recklessid
Hi, I worked out a solution for this problem but I need someone to check it:

Prove that the roots of $\displaystyle (x-a)(x-b) = h^2$ are always real:

$\displaystyle x^2 -bx-ax+ab-h^2 =0$
$\displaystyle A= 1, B= -(a+b), C= ab-h^2$ where $\displaystyle B^2 - 4AC$ = positive or zero for the roots to be real.

$\displaystyle B^2-4AC = [-(a+b)]^2 -4(1)(ab-h^2)$

$\displaystyle a^2+2ab+b^2-4ab+4h^2 = a^2-2ab+b^2 +4h^2$

$\displaystyle = (a-b)^2 + 4h^2$

$\displaystyle (a-b)^2+4h^2$ consists of two perfect squares, which can never be negative. Specifically, the roots are rational because $\displaystyle B^2- 4AC$ is a perfect square.

yes, that works. just say
$\displaystyle (a - b)^2 \ge 0$ and $\displaystyle 4h^2 = (2h)^2 \ge 0$ for all $\displaystyle a,b,h \in \mathbb{R}$ (since they are squares), so that the discriminant $\displaystyle b^2 - 4ac \ge 0$ and we have real roots
• Jun 25th 2008, 02:31 PM
thelostchild
That looks perfect to me nothing I can see wrong with it :)