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Chapter 3: Problem 21

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ 2 x^{2}+3 x>0 $$

Short Answer

Expert verified
The solution for the inequality \(2x^{2} + 3x > 0\) in interval notation is \((-3/2, 0) \cup (0, \infty)\).

Step by step solution

01

Factoring the inequality

We begin by factoring the polynomial inequality to find the roots. The polynomial \(2x^{2} + 3x\) can be written as \(x(2x + 3)\). Therefore, the inequality becomes \(x(2x + 3) > 0\).
02

Solving for x

The next step is to solve for x by setting each of the factors equal to zero. Solving \(x = 0\) gives x = 0 as a solution and for \(2x + 3 = 0\) gives \(x = -3/2\). Now the number line is broken into three intervals by the two numbers 0 and \(-3/2\).
03

Test the intervals

It's important to test each interval to determine if the inequality holds. Choose \(x = -2\) in the first interval \(-\infty, -3/2\), \(x = -1\) in the second interval \(-3/2, 0\), and \(x = 1\) in the third interval \(0, \infty\). Now, substitute these values in the factored inequality \(x(2x + 3) > 0\). The inequality holds for the intervals with positive results.
04

Solution in interval notation

Express the solution in interval notation. From the intervals where the inequality holds, the solution is \((-3/2, 0) \cup (0, \infty)\).
05

Graph on a number line

Graph the solution set on a number line. Draw an open circle at \(x = -3/2\) and \(x = 0\), draw a line toward -\(\infty\) starting from -3/2, but exclude -3/2 itself, and draw a line toward +\(\infty\) starting from 0, but exclude 0.

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