2B. Determine the real number $x$ such that the sequence $\sqrt{x-7}, \sqrt[4]{2 x+19}$, $\sqrt{x+3}$ is a geometric progression.
Solution. Clearly $x \geq 7$. We have $\sqrt[4]{2 x+19}^{2}=\sqrt{x-7} \sqrt{x+3}$, from which we get

$$
2 x+19=(x-7)(x+3), \text { i.e. } x^{2}-6 x-40=0
$$

The solutions to the last quadratic equation are $x_{1}=-4$ and $x_{2}=10$. However, $x \geq 7$, so the only value of $x$ for which we get a geometric progression is $x=10$.