Solve in $\mathbb{R}$ the equation $\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10}=5$.

#
Solution

First Solution: We can eliminate radicals by squaring both sides of the equation below:

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10} & =5 \\
\left(\sqrt{x^{2}-6 x+10}\right)^{2} & =\left(5-\sqrt{x^{2}+9}\right)^{2} \\
x^{2}-6 x+10 & =25-10 \sqrt{x^{2}+9}+x^{2}+9 \\
10 \sqrt{x^{2}+9} & =6 x+24 \\
25\left(x^{2}+9\right) & =(3 x+12)^{2} \\
25 x^{2}+225 & =9 x^{2}+72 x+144 \\
16 x^{2}-72 x+81 & =0 \\
(4 x-9)^{2} & =0
\end{aligned}
$$

Consequently, $4 x-9=0$, or $x=\frac{9}{4}$. To verify that $x=\frac{9}{4}$ is the solution, we can write

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10} & =\sqrt{\frac{81}{16}+9}+\sqrt{\frac{81}{16}-6 \cdot \frac{9}{4}+10} \\
& =\frac{15}{4}+\frac{5}{4} \\
& =5
\end{aligned}
$$

Second Solution: Consider points $F, D$, and $E$ in the Cartesian plane with coordinates $(x, 0)$, $(0,3)$, and $(3,-1)$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-35.jpg?height=597&width=720&top_left_y=1632&top_left_x=768)

We can associate the distances between some points with the given radicals:

$$
\begin{aligned}
D F & =\sqrt{x^{2}+9} \\
F E & =\sqrt{(x-3)^{2}+1^{2}} \\
& =\sqrt{x^{2}-6 x+10} \\
D E & =\sqrt{9+16} \\
& =5
\end{aligned}
$$

By the triangle inequality,

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{(x-3)^{2}+1^{2}} & =D F+F E \\
& \geq D E \\
& =5
\end{aligned}
$$

Equality holds only when $D, F$, and $E$ are collinear. Therefore, the point $F$ must coincide with the intersection $G$ between $D E$ and the $O x$ axis, i.e., $x=9 / 4$.

Observation: A related problem that could be approached similarly using Analytic Geometry is:

What is the minimum value of the real function

$$
f(x)=\sqrt{a^{2}+x^{2}}+\sqrt{(b-x)^{2}+c^{2}}
$$

where $a, b, c$ are positive real numbers?

Repeating the previous argument, if $E=(b,-c), D=(0, a)$, and $F=(x, 0)$, then $f(x)=F E+F D$ and, by the triangle inequality, $f(x) \geq D E$ where equality occurs only if $D, F$, and $E$ are collinear. The intersection of $D E$ and the $O x$ axis is $\left(\frac{a b}{a+c}, 0\right)$, and thus the minimum occurs at $x=\frac{a b}{a+c}$.