Question

x²-（2k+3）x+k²+3k+2=0这方程怎么解?

Answer 1

x²-(2k+3)x+k²+3k+2＝0

→x²-[(k+1)+(k+2)]x+(k+1)(k+2)＝0

→[x-(k+1)][x-(k+2)]＝0

∴x1＝k+1, x2＝k+2.