Problem 5. Solve the inequality $x-2 \leqslant \sqrt{x}$.
![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-077.jpg?height=556&width=1358&top_left_y=258&top_left_x=357)
Solution. We construct the graphs of the functions $f(x)=x-2$ and $\varphi(x)=\sqrt{x}$ (Fig. 4). It then becomes clear that the equation $x-2=\sqrt{x}$ has only one positive root $x_{0}>2$, and the solution to the given inequality is the interval $\left[0 ; x_{0}\right]$, since the inequality is defined only for non-negative numbers.

By solving the equation $x-2=\sqrt{x}$ by squaring both positive parts, we get $x_{0}=4$.

Here, we need to note the following:

Firstly, the equation $(x-2)^{2}=x$ is not equivalent to the equation $x-2=\sqrt{x}$. The first has roots 1 and 4, while the second has only 4. From the graph, it is clear why the number 1 is an extraneous root for the equation $x-2=\sqrt{x}$ (point $A_{1}$ is the intersection of the graphs of the functions $f(x)=x-2$ and $\left.\varphi_{1}(x)=-\sqrt{x}\right)$.

Secondly, the graphs of the functions $f(x)$ and $\varphi(x)$ constructed in advance eliminate the question of extraneous roots of the corresponding equations and inequalities.