sqrt(2+2x)≤4. inequation

Given the inequality:
$$\sqrt{2 x + 2} \leq 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 x + 2} = 4$$
Solve:
Given the equation
$$\sqrt{2 x + 2} = 4$$
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:
$$\left(\sqrt{2 x + 2}\right)^{2} = 4^{2}$$
or
$$2 x + 2 = 16$$
Move free summands (without x)
from left part to right part, we given:
$$2 x = 14$$
Divide both parts of the equation by 2

x = 14 / (2)

We get the answer: x = 7

$$x_{1} = 7$$
$$x_{1} = 7$$
This roots
$$x_{1} = 7$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 7$$
=
$$\frac{69}{10}$$
substitute to the expression
$$\sqrt{2 x + 2} \leq 4$$
$$\sqrt{2 + \frac{2 \cdot 69}{10}} \leq 4$$

  _____     
\/ 395      
------- <= 4
   5        
     

the solution of our inequality is:
$$x \leq 7$$

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       x1