已知x1,x2是关于x的方程x²+(2m-1)x+m²+1=0的两个实数根,当x1²+x2²=15时,求m的值

x1+x2=-2m+1x1*x2=m^2+1x1^2+x2^2=(x1+x2)^2-2x1x2=(-2m+1)^2 -2(m^2+1)=4m^2-4m+1-2m^2-2=2m^2-4m-1=15得2m^2-4m-16=0m^2-2m-8=0(m-1)^2=9m1=4m2 =-2又(2m-1)^2-4(m^2+1)=4m^2-4m+1-4m^2-4=-4m-3>=0m