Question

已知关于x的一元二次方程①（k+2)x²+x+2=0和②（k+2)x²+kx+k+1=0

，（3）若方程②的两根之差为1时，求k的值.

Answer 1

②（k+2)x²+kx+k+1=0
x1+x2=-k/(k+2)
x1*x2=(k+1)/(k+2)
x1-x2=1

(x1+x2)²=k²/(k+2)²=x1²+x2²+2x1x2
x1²+x2²=k²/(k+2)²-2(k+1)/(k+2)
(x1-x2)²=1=x1²+x2²-2x1x2=k²/(k+2)²-2(k+1)/(k+2)-2(k+1)/(k+2)
k²+4k+3=0
k=-1 k=-3