sqrt(x)<=√2/2 inequation

Given the inequality:
$$\sqrt{x} \leq \frac{\sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{x} = \frac{\sqrt{2}}{2}$$
Solve:
Given the equation
$$\sqrt{x} = \frac{\sqrt{2}}{2}$$
Because equation degree is equal to = 1/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2-th degree:
We get:
$$\left(\sqrt{x}\right)^{2} = \left(\frac{\sqrt{2}}{2}\right)^{2}$$
or
$$x = \frac{1}{2}$$
We get the answer: x = 1/2

$$x_{1} = \frac{1}{2}$$
$$x_{1} = \frac{1}{2}$$
This roots
$$x_{1} = \frac{1}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{2}$$
=
$$\frac{2}{5}$$
substitute to the expression
$$\sqrt{x} \leq \frac{\sqrt{2}}{2}$$
$$\sqrt{\frac{2}{5}} \leq \frac{\sqrt{2}}{2}$$

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\/ 10     \/ 2 
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  5         2  
    

the solution of our inequality is:
$$x \leq \frac{1}{2}$$

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       x1