Given x_1,x_2 are the real roots of x^2-2mx+(m^2+4m+5)=0, please determine the minimum value of (x_1)^2+(x_2)^2.
+118608 x^2-2mx+(m^2+4m+5)=0
\(x^2-2mx+(m^2+4m+5)=0\\ x=\frac{2m\pm \sqrt{4m^2-4(m^2+4m+5)}}{2}\\ x=m\pm \sqrt{m^2-m^2+4m+5}\\ x=m\pm \sqrt{4m+5}\\ \mbox{real roots when }\\ 4m+5\ge0\\ m\ge-1.25 \)
Not finished, someone else is welcome to continue if they want to.
Coding:
x^2-2mx+(m^2+4m+5)=0\\
x=\frac{2m\pm \sqrt{4m^2-4(m^2+4m+5)}}{2}\\
x=m\pm \sqrt{m^2-m^2+4m+5}\\
x=m\pm \sqrt{4m+5}\\
\mbox{real roots when }\\
4m+5\ge0\\
m\ge-1.25
