X1、X2為關于4X方-（3M-5）X-6M方=0的兩實根，且X1除以X2的絕對值=3/2，求M的值

熱心網友

由根與系數的關系：X1+X2=(3M-5)/4,X1*X2=(-6M^2)/4=-(3M^2)/2 == X1與X2異號∴X1/X2=-3/2 == X2=-(2/3)X1代入上面兩式，得到X1=(9M-15)/4,M^2=(4/9)X1^2== M^2=[(9M-15)^2]/16 == M^2-6M+5=0 == M=1或M=5

熱心網友

由韋達定理得X1+X2=(3M-5)/4(1)X1*X2=-3M/2|x1/X2|=3/2X1/X2=-3/2X1=-3X2/2(1)^2 X1^2+X2^2+2X1*X2=M+9M/4-3M=(3M-5)^2/16M=1或25/9不知有沒有算錯。