sqrt(3*x+1)<1 inequation

Given the inequality:
$$\sqrt{3 x + 1} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{3 x + 1} = 1$$
Solve:
Given the equation
$$\sqrt{3 x + 1} = 1$$
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:
$$\left(\sqrt{3 x + 1}\right)^{2} = 1^{2}$$
or
$$3 x + 1 = 1$$
Move free summands (without x)
from left part to right part, we given:
$$3 x = 0$$
Divide both parts of the equation by 3

x = 0 / (3)

We get the answer: x = 0

$$x_{1} = 0$$
$$x_{1} = 0$$
This roots
$$x_{1} = 0$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$\sqrt{3 x + 1} < 1$$
$$\sqrt{\frac{\left(-1\right) 3}{10} + 1} < 1$$

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\/ 70     
------ < 1
  10      
    

the solution of our inequality is:
$$x < 0$$

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       x1