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**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.