ax²+bx+c=0 ∵方程有兩個(gè)實(shí)數(shù)根∴Δ＞0
x1=(－b＋√b²－4ac)／2a, x2=(－b-√b²－4ac)／2a
x1²=(b²-2b√b²－4ac+(b²－4ac))/4a²
x2²=(b²+2b√b²－4ac+(b²－4ac))/4a²
∴x1²+x2²=(b²-2ac)/a²∵k≥3
((-2(k-1))²-2(k-2)(k+1))/(k-2)²=4
4k²-8k+4-2k²+2k+4=4k²-16k+16
2k²-10k+8=0
(k-1)(k-4)=0∴k=4