Aufgabe:
(x-a+b)(x-a+c) = (a-b)2 - x2
Lösung: a-b; \( \frac{b-c}{2} \)
Problem/Ansatz:
x2-ax+bx-ax+a2-ab+cx-ac+bc = a2-2ab+b2-x2
2x2-2ax+bx+cx+ab-ac+bc-b2 = 0
2x2-x(2a-b-c) + a(b-c) + b(c-b) = 0
2x2-x(2a-b-c) + a(b-c) - b(b-c) = 0
2x2-x(2a-b-c) + (a-b)(b-c) = 0
x2-x\( \frac{2a-b-c}{2} \) + \( \frac{(a-b)(b-c}{2} \)
x1;2 = \( \frac{2a-b-c}{4} \) ± \( \sqrt{(\frac{2a-b-c}{4})^2-\frac{(a-b)(b-c)}{2}} \)
wie geht es weiter?