Given the inequality:
$$64 x^{2} - 36 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$64 x^{2} - 36 = 0$$
Solve:
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
$$a = 64$$
$$b = 0$$
$$c = -36$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (64) * (-36) = 9216
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{3}{4}$$
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = \frac{3}{4}$$
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = \frac{3}{4}$$
$$x_{2} = - \frac{3}{4}$$
This roots
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = \frac{3}{4}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{3}{4} + - \frac{1}{10}$$
=
$$- \frac{17}{20}$$
substitute to the expression
$$64 x^{2} - 36 > 0$$
$$-36 + 64 \left(- \frac{17}{20}\right)^{2} > 0$$
256
--- > 0
25
one of the solutions of our inequality is:
$$x < - \frac{3}{4}$$
_____ _____
\ /
-------ο-------ο-------
x2 x1
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{3}{4}$$
$$x > \frac{3}{4}$$