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(x+12)*(x-5)›0 inequation

A inequation with variable

The solution

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(x + 12)*(x - 5) > 0
$$\left(x - 5\right) \left(x + 12\right) > 0$$
(x - 5)*(x + 12) > 0
Detail solution
Given the inequality:
$$\left(x - 5\right) \left(x + 12\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 5\right) \left(x + 12\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - 5\right) \left(x + 12\right) = 0$$
We get the quadratic equation
$$x^{2} + 7 x - 60 = 0$$
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
$$a = 1$$
$$b = 7$$
$$c = -60$$
, then
D = b^2 - 4 * a * c = 

(7)^2 - 4 * (1) * (-60) = 289

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

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

or
$$x_{1} = 5$$
$$x_{2} = -12$$
$$x_{1} = 5$$
$$x_{2} = -12$$
$$x_{1} = 5$$
$$x_{2} = -12$$
This roots
$$x_{2} = -12$$
$$x_{1} = 5$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-12 + - \frac{1}{10}$$
=
$$- \frac{121}{10}$$
substitute to the expression
$$\left(x - 5\right) \left(x + 12\right) > 0$$
$$\left(- \frac{121}{10} - 5\right) \left(- \frac{121}{10} + 12\right) > 0$$
171    
--- > 0
100    

one of the solutions of our inequality is:
$$x < -12$$
 _____           _____          
      \         /
-------ο-------ο-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -12$$
$$x > 5$$
Solving inequality on a graph
Rapid solution [src]
Or(And(-oo < x, x < -12), And(5 < x, x < oo))
$$\left(-\infty < x \wedge x < -12\right) \vee \left(5 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -12))∨((5 < x)∧(x < oo))
Rapid solution 2 [src]
(-oo, -12) U (5, oo)
$$x\ in\ \left(-\infty, -12\right) \cup \left(5, \infty\right)$$
x in Union(Interval.open(-oo, -12), Interval.open(5, oo))
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