B2. If the square root of a number is divided by $\frac{1}{2}$, we get exactly as much as if the number is reduced by 3. What is this number? Are there more such numbers?
B2. The equation $\sqrt{x}: \frac{1}{2}=x-3$ is transformed into $x^{2}-10 x+9=0$, which has solutions $x_{1}=9$ and $x_{2}=1$. The number 9 satisfies the condition of the problem, as $\sqrt{9}: \frac{1}{2}=9-3$, while the number 1 does not, since $\sqrt{1}: \frac{1}{2} \neq 1-3$.

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=62&width=1639&top_left_y=1371&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-12.jpg?height=51&width=1639&top_left_y=1422&top_left_x=274)

Solutions to the quadratic equation $x^{2}-10 x+9=0 ; x_{1}=9, x_{2}=1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Verification.................................................................................................................................................

Answer: The sought number is 9. ......................................................................................