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25x^2>_64 inequation

A inequation with variable

The solution

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    2      
25*x  >= 64
$$25 x^{2} \geq 64$$
25*x^2 >= 64
Detail solution
Given the inequality:
$$25 x^{2} \geq 64$$
To solve this inequality, we must first solve the corresponding equation:
$$25 x^{2} = 64$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$25 x^{2} = 64$$
to
$$25 x^{2} - 64 = 0$$
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
$$a = 25$$
$$b = 0$$
$$c = -64$$
, then
D = b^2 - 4 * a * c = 

(0)^2 - 4 * (25) * (-64) = 6400

Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = \frac{8}{5}$$
$$x_{2} = - \frac{8}{5}$$
$$x_{1} = \frac{8}{5}$$
$$x_{2} = - \frac{8}{5}$$
$$x_{1} = \frac{8}{5}$$
$$x_{2} = - \frac{8}{5}$$
This roots
$$x_{2} = - \frac{8}{5}$$
$$x_{1} = \frac{8}{5}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{8}{5} + - \frac{1}{10}$$
=
$$- \frac{17}{10}$$
substitute to the expression
$$25 x^{2} \geq 64$$
$$25 \left(- \frac{17}{10}\right)^{2} \geq 64$$
289/4 >= 64

one of the solutions of our inequality is:
$$x \leq - \frac{8}{5}$$
 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{8}{5}$$
$$x \geq \frac{8}{5}$$
Solving inequality on a graph
Rapid solution [src]
Or(And(8/5 <= x, x < oo), And(x <= -8/5, -oo < x))
$$\left(\frac{8}{5} \leq x \wedge x < \infty\right) \vee \left(x \leq - \frac{8}{5} \wedge -\infty < x\right)$$
((8/5 <= x)∧(x < oo))∨((x <= -8/5)∧(-oo < x))
Rapid solution 2 [src]
(-oo, -8/5] U [8/5, oo)
$$x\ in\ \left(-\infty, - \frac{8}{5}\right] \cup \left[\frac{8}{5}, \infty\right)$$
x in Union(Interval(-oo, -8/5), Interval(8/5, oo))