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Solve for $x$ . $\newline$ $(x - 2)(x + 3) \geq 0$ $\newline$ Write a compound inequality like $1 < x < 3$ or like $x < 1$ or $x > 3$ .

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Solve quadratic inequalities

Full solution

Q. Solve for

x

.

\newline

(x - 2)(x + 3) \geq 0

\newline

Write a compound inequality like

1 < x < 3

or like

x < 1

or

x > 3

.

Find Zeros: Find the zeros of the inequality by setting $(x - 2)(x + 3)$ equal to $0$ . $x - 2 = 0$ or $x + 3 = 0$ $x = 2$ or $x = -3$
Determine Intervals: Determine the intervals to test based on the zeros: $(-\infty, -3)$ , $(-3, 2)$ , $(2, \infty)$ .
Test Values: Test a value from the interval $(-\infty, -3)$ , say $x = -4$ . $\newline$ $(-4 - 2)(-4 + 3) = (-6)(-1) = 6$ , which is positive.
Combine Intervals: Test a value from the interval $-3, 2$ , say $x = 0$ . $(0 - 2)(0 + 3) = (-2)(3) = -6$ , which is negative.
Combine Intervals: Test a value from the interval $(-3, 2)$ , say $x = 0$ . $\newline$ $(0 - 2)(0 + 3) = (-2)(3) = -6$ , which is negative.Test a value from the interval $(2, \infty)$ , say $x = 4$ . $\newline$ $(4 - 2)(4 + 3) = (2)(7) = 14$ , which is positive.
Combine Intervals: Test a value from the interval $-3, 2$ , say $x = 0$ . $(0 - 2)(0 + 3) = (-2)(3) = -6$ , which is negative.Test a value from the interval $(2, \infty)$ , say $x = 4$ . $(4 - 2)(4 + 3) = (2)(7) = 14$ , which is positive.Combine the intervals where the inequality $(x - 2)(x + 3)$ is non-negative. The solution is $x \in (-\infty, -3] \cup [2, \infty)$ .

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1

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