(1-4x)^2-(8x-1)(2x+1)>0 inequation

Given the inequality:
$$\left(1 - 4 x\right)^{2} - \left(2 x + 1\right) \left(8 x - 1\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(1 - 4 x\right)^{2} - \left(2 x + 1\right) \left(8 x - 1\right) = 0$$
Solve:
Given the equation:

(1-4*x)^2-(8*x-1)*(2*x+1) = 0

Expand expressions:

1 - 8*x + 16*x^2 - (8*x - 1)*(2*x + 1) = 0

1 - 8*x + 16*x^2 + 1 - 16*x^2 - 6*x = 0

Reducing, you get:

2 - 14*x = 0

Move free summands (without x)
from left part to right part, we given:
$$- 14 x = -2$$
Divide both parts of the equation by -14

x = -2 / (-14)

We get the answer: x = 1/7
$$x_{1} = \frac{1}{7}$$
$$x_{1} = \frac{1}{7}$$
This roots
$$x_{1} = \frac{1}{7}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{7}$$
=
$$\frac{3}{70}$$
substitute to the expression
$$\left(1 - 4 x\right)^{2} - \left(2 x + 1\right) \left(8 x - 1\right) > 0$$
$$\left(1 - \frac{3 \cdot 4}{70}\right)^{2} - \left(-1 + \frac{3 \cdot 8}{70}\right) \left(\frac{2 \cdot 3}{70} + 1\right) > 0$$

7/5 > 0

the solution of our inequality is:
$$x < \frac{1}{7}$$

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