Question

设x1,x2为方程4x^2-4mx+m+2=0的两个实数根,当m为何值时,x1^2+x2^2的值最小?求出这个最小值

Answer 1

x1,x2为方程4x^2-4mx+m+2=0的两个实数根

x1+x2=4m/4=m

x1x2=(m+2)/4

x1^2+x2^2=(x1+x2)^2-2x1x2

=m^2-(m+2)/2

=m^2-m/2+1

=(m-1/4)^2+3/4

所以,m=1/4时,有最小值=3/4