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Chapter 3: Problem 20

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. $$ x^{2}+2 x<0 $$

Short Answer

Expert verified
The solution set for the inequality \(x^2 + 2x < 0\) is \((-2, 0)\).

Step by step solution

01

Solve the Quadratic Equation

First, solve the equation \(x^2 + 2x = 0\). This can be rewritten as \(x(x + 2) = 0\). It leads to two roots: \(x_1 = 0\) and \(x_2 = -2\) when each factor is independently set to zero.
02

Determine the Intervals

With these two roots, we can divide the number line into three intervals: \(-\infty , -2\), \(-2 , 0\), and \(0 , +\infty\). We pick a test number from each interval to determine the sign of the quadratic function. Let's take these testing points -3 for the first interval, -1 for the second interval, and 1 for the last interval.
03

Test the Intervals

Substituting -3 into the inequality \(x^2 + 2x < 0\), we get \((-3)^2 + 2*(-3) = 9 - 6 = 3\) which is greater than 0. Thus all values in the interval \(-\infty , -2\) are not part of the solution set. Substituting -1, we get \((-1)^2 + 2*(-1) = 1 - 2 = -1\) which is less than 0, thus all values in the interval \(-2 , 0\) are part of the solution set. Substituting 1, we get \((1)^2 + 2*1 = 1 + 2 = 3\) which is greater than 0. Thus all values in the interval \(0 , +\infty\) are not part of the solution set.
04

Write the Solution in Interval Notation

The solution set of the inequality includes only those values of x for which the inequality holds true. The inequality holds true only in the interval \(-2 , 0\). Therefore, the solution in interval notation will be \((-2, 0)\).

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