<s>[INST] Real solutions of the equation $x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x}$ Solve the equation
$$x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x},$$
where $x \in \mathbb{R}$.
 [/INST] I'm assuming we are solving over the reals.
Set $a=\sqrt{3-x}$ and likewise for $b$ and $c$.
Then $x=ab+bc+ca$ and we see $$a^2+x = 3,\qquad b^2+x=4,\qquad c^2+x=5$$
Substituting for $x$, we find $$(a+b)(a+c) = 3,\qquad (b+c)(b+a)=4,\qquad (c+a)(c+b)=5$$
Thus $$(a+b)(a+c)(b+c)=\sqrt{60}$$
This leads to $$a+b=\frac{\sqrt{60}}{5},\qquad b+c=\frac{\sqrt{60}}{3},\qquad a+c=\frac{\sqrt{60}}{4}$$
So $$a = a+b+c - (b+c) = \frac{47}{2\sqrt{60}} - \frac{20}{\sqrt{60}} = \frac{7}{2\sqrt{60}}$$
Thus $$x = 3-a^2 = 3 - \frac{7^2}{240} = \boxed{\frac{671}{240}}$$
</s>