sqrt2x-3<=0 inequation

Given the inequality:
$$\sqrt{2 x} - 3 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 x} - 3 = 0$$
Solve:
Given the equation
$$\sqrt{2 x} - 3 = 0$$
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:
$$\left(\sqrt{2 x}\right)^{2} = 3^{2}$$
or
$$2 x = 9$$
Divide both parts of the equation by 2

x = 9 / (2)

We get the answer: x = 9/2

$$x_{1} = \frac{9}{2}$$
$$x_{1} = \frac{9}{2}$$
This roots
$$x_{1} = \frac{9}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{9}{2}$$
=
$$\frac{22}{5}$$
substitute to the expression
$$\sqrt{2 x} - 3 \leq 0$$
$$-3 + \sqrt{\frac{2 \cdot 22}{5}} \leq 0$$

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     2*\/ 55      
-3 + -------- <= 0
        5         
     

the solution of our inequality is:
$$x \leq \frac{9}{2}$$

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