sqrt(6x-9)>x inequation

Given the inequality:
$$\sqrt{6 x - 9} > x$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{6 x - 9} = x$$
Solve:
Given the equation
$$\sqrt{6 x - 9} = x$$
$$\sqrt{6 x - 9} = x$$
We raise the equation sides to 2-th degree
$$6 x - 9 = x^{2}$$
$$6 x - 9 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 6 x - 9 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = -1$$
$$b = 6$$
$$c = -9$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) \left(\left(-1\right) 4\right) \left(-9\right) + 6^{2} = 0$$
Because D = 0, then the equation has one root.

x = -b/2a = -6/2/(-1)

$$x_{1} = 3$$

Because
$$\sqrt{6 x - 9} = x$$
and
$$\sqrt{6 x - 9} \geq 0$$
then
$$x >= 0$$
or
$$0 \leq x$$
$$x < \infty$$
$$x_{1} = 3$$
$$x_{1} = 3$$
$$x_{1} = 3$$
This roots
$$x_{1} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 3$$
=
$$\frac{29}{10}$$
substitute to the expression
$$\sqrt{6 x - 9} > x$$
$$\sqrt{\left(-1\right) 9 + 6 \cdot \frac{29}{10}} > \frac{29}{10}$$

  _____     
\/ 210    29
------- > --
   5      10
     

Then
$$x < 3$$
no execute
the solution of our inequality is:
$$x > 3$$

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        /
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       x_1