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Chapter 2: Problem 21

Solve each polynomial inequality 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 of the polynomial inequality \(2x^2 + 3x > 0\) is \((-鈭�, -1.5)\) Union \((0, +鈭�)\), and the graph of the solution set on a real number line is two shaded intervals: -鈭� to -1.5 and 0 to 鈭�, with open circles at -1.5 and 0.

Step by step solution

01

Simplify the inequality

The first step is to factor the polynomial. To do this, factor out an \(x\) from each term. So, \(2x^2 + 3x > 0\) simplifies to \(x(2x + 3) > 0\). Now, the inequality is in a form that we can easily work with.
02

Determine the intervals

To find the solution set, set the factors of the inequality to 0 and solve for \(x\). This gives the possible points where the polynomial changes sign. Solve \(x = 0\) and \(2x + 3 = 0\) which gives \(x = -1.5\) and \(x = 0\). These two solutions divide the number line into three intervals: \((-鈭�, -1.5)\), \((-1.5, 0)\), and \((0, +鈭�)\). Now, substitute a number that falls within each interval into the inequality to determine if that interval is part of the solution.
03

Test the intervals

Choose representative numbers for each interval. The numbers -2, -1, and 1 are good choices. Substituting the representative values into \(x(2x + 3)\) inequality: When \(x = -2\), the expression yields a positive value; when \(x = -1\) and \(x = 1\), it yields a negative value. The original inequality is \(x(2x + 3) > 0\), indicates that x should produce a positive result. So the solution of the inequality is: \((-鈭�, -1.5)\) Union \((0, +鈭�)\).
04

Graph the solution set

On the number line, we graph the solution set by shading all the values from -鈭� to -1.5 and from 0 to 鈭�. The endpoints -1.5 and 0 are not included in the solution, so open circles are used on these points.

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Most popular questions from this chapter

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