Mister Exam
(x-4)(x+5)/x^2>0 inequation
The solution
You have entered
[src]
(x - 4)*(x + 5)
--------------- > 0
2
x
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} > 0$$
((x - 4)*(x + 5))/x^2 > 0
Detail solution
Given the inequality:
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} = 0$$
Solve:
Given the equation:
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} = 0$$
the denominator
$$x$$
then
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 4 = 0$$
$$x + 5 = 0$$
solve the resulting equation:
2.
$$x - 4 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 4$$
We get the answer: x1 = 4
3.
$$x + 5 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -5$$
We get the answer: x2 = -5
but
$$x_{1} = 4$$
$$x_{2} = -5$$
$$x_{1} = 4$$
$$x_{2} = -5$$
This roots
$$x_{2} = -5$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} > 0$$
$$\frac{\left(- \frac{51}{10} - 4\right) \left(- \frac{51}{10} + 5\right)}{\left(- \frac{51}{10}\right)^{2}} > 0$$
one of the solutions of our inequality is:
$$x < -5$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -5$$
$$x > 4$$
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} = 0$$
Solve:
Given the equation:
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} = 0$$
the denominator
$$x$$
then
x is not equal to 0
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 4 = 0$$
$$x + 5 = 0$$
solve the resulting equation:
2.
$$x - 4 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 4$$
We get the answer: x1 = 4
3.
$$x + 5 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -5$$
We get the answer: x2 = -5
but
x is not equal to 0
$$x_{1} = 4$$
$$x_{2} = -5$$
$$x_{1} = 4$$
$$x_{2} = -5$$
This roots
$$x_{2} = -5$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\frac{\left(x - 4\right) \left(x + 5\right)}{x^{2}} > 0$$
$$\frac{\left(- \frac{51}{10} - 4\right) \left(- \frac{51}{10} + 5\right)}{\left(- \frac{51}{10}\right)^{2}} > 0$$
91 ---- > 0 2601
one of the solutions of our inequality is:
$$x < -5$$
_____ _____
\ /
-------ο-------ο-------
x2 x1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -5$$
$$x > 4$$
Solving inequality on a graph
Rapid solution
[src]
Or(And(-oo < x, x < -5), And(4 < x, x < oo))
$$\left(-\infty < x \wedge x < -5\right) \vee \left(4 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -5))∨((4 < x)∧(x < oo))
Rapid solution 2
[src]
(-oo, -5) U (4, oo)
$$x\ in\ \left(-\infty, -5\right) \cup \left(4, \infty\right)$$
x in Union(Interval.open(-oo, -5), Interval.open(4, oo))