Given the inequality:
$$6 x - \frac{2}{x - 1} \left(x + 2\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$6 x - \frac{2}{x - 1} \left(x + 2\right) = 0$$
Solve:
Given the equation:
$$6 x - \frac{2}{x - 1} \left(x + 2\right) = 0$$
Multiply the equation sides by the denominators:
-1 + x
we get:
$$\left(x - 1\right) \left(6 x - \frac{2}{x - 1} \left(x + 2\right)\right) = 0$$
$$6 x^{2} - 8 x - 4 = 0$$
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
$$a = 6$$
$$b = -8$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(-8)^2 - 4 * (6) * (-4) = 160
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{2}{3} + \frac{\sqrt{10}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{\sqrt{10}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{\sqrt{10}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{\sqrt{10}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{\sqrt{10}}{3}$$
$$x_{2} = \frac{2}{3} - \frac{\sqrt{10}}{3}$$
This roots
$$x_{2} = \frac{2}{3} - \frac{\sqrt{10}}{3}$$
$$x_{1} = \frac{2}{3} + \frac{\sqrt{10}}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$\left(\frac{2}{3} - \frac{\sqrt{10}}{3}\right) + - \frac{1}{10}$$
=
$$\frac{17}{30} - \frac{\sqrt{10}}{3}$$
substitute to the expression
$$6 x - \frac{2}{x - 1} \left(x + 2\right) \leq 0$$
$$6 \left(\frac{17}{30} - \frac{\sqrt{10}}{3}\right) - \frac{2}{-1 + \left(\frac{17}{30} - \frac{\sqrt{10}}{3}\right)} \left(\left(\frac{17}{30} - \frac{\sqrt{10}}{3}\right) + 2\right) \leq 0$$
/ ____\
|77 \/ 10 |
2*|-- - ------|
17 ____ \30 3 /
-- - 2*\/ 10 - --------------- <= 0
5 ____
13 \/ 10
- -- - ------
30 3
one of the solutions of our inequality is:
$$x \leq \frac{2}{3} - \frac{\sqrt{10}}{3}$$
_____ _____
\ /
-------•-------•-------
x2 x1
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{2}{3} - \frac{\sqrt{10}}{3}$$
$$x \geq \frac{2}{3} + \frac{\sqrt{10}}{3}$$