Question

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+x2

Answer 1

x1+x2=-2m+1

x1*x2=m^2+1

x1^2+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=0

m^2-2m-8=0 (m-1)^2=9

m1=4 m2 =-2

又(2m-1)^2-4(m^2+1)=4m^2-4m+1-4m^2-4=-4m-3>=0

m