Given the inequality:
$$\left(x - 3\right) \left(x + 2\right) < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 3\right) \left(x + 2\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x - 3\right) \left(x + 2\right) = 0$$
We get the quadratic equation
$$x^{2} - x - 6 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -6$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-6) = 25
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3$$
$$x_{2} = -2$$
$$x_{1} = 3$$
$$x_{2} = -2$$
$$x_{1} = 3$$
$$x_{2} = -2$$
This roots
$$x_{2} = -2$$
$$x_{1} = 3$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\left(x - 3\right) \left(x + 2\right) < 0$$
$$\left(-3 + - \frac{21}{10}\right) \left(- \frac{21}{10} + 2\right) < 0$$
51
--- < 0
100
but
51
--- > 0
100
Then
$$x < -2$$
no execute
one of the solutions of our inequality is:
$$x > -2 \wedge x < 3$$
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