8. If $x$ is a real number, find the smallest value of $\sqrt{x^{2}+4 x+5}+\sqrt{x^{2}-8 x+25}$.
(1 mark)若 $x$ 是實數, 求 $\sqrt{x^{2}+4 x+5}+\sqrt{x^{2}-8 x+25}$ 的最小值。
8. $2 \sqrt{13}$
8. Rewrite $\sqrt{x^{2}+4 x+5}+\sqrt{x^{2}-8 x+25}$ into the form $\sqrt{(x+2)^{2}+(0-1)^{2}}+\sqrt{(x-4)^{2}+(0+3)^{2}}$. If we consider the three points $A(-2,1), B(x, 0)$ and $P(4,-3)$, then the first term is the distance between $A$ and $P$ while the second is the distance between $P$ and $B$. The minimum of the sum this occurs when $A, P, B$ are collinear, and the minimum sum is equal to the distance between $A$ and $B$, which is $\sqrt{(-2-4)^{2}+[1-(-3)]^{2}}=2 \sqrt{13}$.