sqrt(3x+1)>3x inequation

Given the inequality:
$$\sqrt{3 x + 1} > 3 x$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{3 x + 1} = 3 x$$
Solve:
Given the equation
$$\sqrt{3 x + 1} = 3 x$$
$$\sqrt{3 x + 1} = 3 x$$
We raise the equation sides to 2-th degree
$$3 x + 1 = 9 x^{2}$$
$$3 x + 1 = 9 x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- 9 x^{2} + 3 x + 1 = 0$$
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -9$$
$$b = 3$$
$$c = 1$$
, then

D = b^2 - 4 * a * c = 

(3)^2 - 4 * (-9) * (1) = 45

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = \frac{1}{6} - \frac{\sqrt{5}}{6}$$
Simplify
$$x_{2} = \frac{1}{6} + \frac{\sqrt{5}}{6}$$
Simplify

Because
$$\sqrt{3 x + 1} = 3 x$$
and
$$\sqrt{3 x + 1} \geq 0$$
then
$$3 x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
$$x_{2} = \frac{1}{6} + \frac{\sqrt{5}}{6}$$
$$x_{1} = \frac{1}{6} + \frac{\sqrt{5}}{6}$$
$$x_{1} = \frac{1}{6} + \frac{\sqrt{5}}{6}$$
This roots
$$x_{1} = \frac{1}{6} + \frac{\sqrt{5}}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{1}{6} + \frac{\sqrt{5}}{6}\right)$$
=
$$\frac{1}{15} + \frac{\sqrt{5}}{6}$$
substitute to the expression
$$\sqrt{3 x + 1} > 3 x$$
$$\sqrt{1 + 3 \cdot \left(\frac{1}{15} + \frac{\sqrt{5}}{6}\right)} > 3 \cdot \left(\frac{1}{15} + \frac{\sqrt{5}}{6}\right)$$

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the solution of our inequality is:
$$x < \frac{1}{6} + \frac{\sqrt{5}}{6}$$

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