<s>[INST] How did root of the equation $x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$ vanish while solving? The original question was to solve for $x \in \mathbb{R}$ in:
$$x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$$
My Solution: 
By domain requirements: $x \in (-\infty,2) \cup (5,\infty) $.
\begin{align}
x &= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)} \\
\Rightarrow (x - \sqrt{(2-x)(3-x)})^2 &= (\sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)})^2 \\
\Rightarrow 2x^2 - 5x + 6 - 2x\sqrt{(2-x)(3-x)} &= 2x^2 - 15x + 25 + 2(5-x)\sqrt{(2-x)(3-x)} \\
\Rightarrow x - 1.9 &= \sqrt{(3-x)(2-x)} \\
\Rightarrow x \geq 1.9 &\hspace{0.5 cm}\&\hspace{0.5 cm}  (x - 1.9)^2 = (3-x)(2-x) \\
\Rightarrow x = \frac{239}{120}
\end{align}
Question: If I check for the answers of the original equation in WOLFRAM, I get two possible answers: $x = \frac{239}{120} $ and $x = \frac{1}{2}(5 + \sqrt{5} + \sqrt{8 + 3\sqrt{5}})$. My question is where (and why) in my solution did the second root vanish? I have only used squaring and simplification, and as far as I know squaring only creates an additional root, it does not destroy a root.
Link to Wolfram's solution: http://www.wolframalpha.com/input/?i=x%3D+%5Csqrt%7B(2-x)(3-x)%7D+%2B+%5Csqrt%7B(3-x)(5-x)%7D+%2B+%5Csqrt%7B(2-x)(5-x)%7D
 [/INST] Your error occurs at the step where you removed the factor $(5-x)$.

It should be $|5-x|$, which leads to two cases.

If $x \le 5$, then $(5-x)$ is correct.

If $x > 5$, the correct factor is $(x-5)$.
</s>