Question

(-b+√b²-4ac/2a)²+(-b-√b²-4ac/2a)²

(b²-4ac≥0.a≠0)

Answer 1

{[-b+√(b²-4ac)]/(2a)}²+{[-b-√(b²-4ac)/(2a)}²?
b²-4ac≥0.a≠0,
设x1，x2为方程ax²+bx+c=0的两个实数根，
则x1={[-b+√(b²-4ac)]/(2a)}，x2={[-b-√(b²-4ac)/(2a)}，
由韦达定理：x1+x2=-b/a，x1x2=c/a；
原式=x1²+x2²
=（x1+x2）²-2x1x2
=(-b/a)²-2c/a
=b²/a²-2c/a
=(b²-2ac)/a²

Answer 2

(-b+√b²-4ac/2a)²+(-b-√b²-4ac/2a)²
=(-b)²-2b*√(b²-4ac)/2a+[√(b²-4ac)/2a]²+(-b)²+2b*√(b²-4ac)/2a+[√(b²-4ac)/2a]²
=2b²+2(b²-4ac)/4a²
=2b²+(2b²-8ac)/4a²
=(8a²b²+2b²-8ac)/4a²
=(4a²b²+b²-4ac)/2a²