Solve for x. (x - 2)(x + 3) ≥ 0 Write a compound inequality like 1 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 = -3Determine Intervals: Determine the intervals to test based on the zeros: (-∞, -3), (-3, 2), (2, ∞).Test Values: Test a value from the interval (-∞, -3), say x = -4. (-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.Test a value from the interval (2, ∞), 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 ∈ (-∞, -3] ∪ [2, ∞).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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.

View step-by-step help

Home

Math Problems

Algebra 2

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

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 $x \in (-\infty, -3] \cup [2, \infty)$ .