Question

假设a,b是方程的x^2-2mx+1-m^2=0得两个实根,则a^2+b^2得最小值是?

Answer 1

a+b=2m ab=1-m^2

a^2+b^2＝（a+b）^2-2ab=4m^2-2(1-m^2)=2m^2-2

m^2>=0

a^2+b^2＝（a+b）^2-2ab=4m^2-2(1-m^2)=2m^2-2>=-2