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# Help algebra

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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.

Jun 5, 2021

#1
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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

Jun 5, 2021