6.031. $\sqrt{3 x+4}+\sqrt{x-4}=2 \sqrt{x}$.
## Solution.

Domain of definition: $3 x+4 \geq 0, x-4 \geq 0, x \geq 0 \Rightarrow x \geq 4$.

Squaring both sides of the equation, we get

$$
\begin{aligned}
& 3 x+4+2 \sqrt{(3 x+4)(x-4)}+x-4=4 x \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{(3 x+4)(x-4)}=0
\end{aligned}
$$

Squaring again, we get: $(3 x+4)(x-4)=0$. From this, we have $3 x+4=0$ or $x-4=0, x_{1}=-\frac{4}{3}, x_{2}=4 ; x_{1}=-\frac{4}{3}$ does not satisfy the domain of definition. Checking $x=4$ by direct substitution into the original equation, we get:

Answer: $x=4$.