Page 239

messersmith_power_intermediate_algebra_1e_ch4_7_10

in the interval makes x2 2x 3 positive, then all numbers in that interval will make x2 2x 3 positive.) Indicate the result on the number line. Interval B Negative Positive 1 3 Interval A Positive Interval C x2 2x 3 Interval A: (x 1) As a test number, choose any number less than 1. We will choose 2. Evaluate x2 2x 3 for x 2. x2 2x 3 (2)2 2(2) 3 4 4 3 8 3 5 Substitute 2 for x. When x 2, x2 2x 3 is positive. Therefore, x2 2x 3 will be positive for all values of x in this interval. Indicate this on the number line as seen above. Interval B: (1 x 3) As a test number, choose any number between 1 and 3. We will choose 0. Evaluate x2 2x 3 for x 0. x2 2x 3 (0)2 2(0) 3 0 0 3 3 Substitute 0 for x. When x 0, x2 2x 3 is negative. Therefore, x2 2x 3 will be negative for all values of x in this interval. Indicate this on the number line above. Interval C: (x 3) As a test number, choose any number greater than 3. We will choose 4. Evaluate x2 2x 3 for x 4. x2 2x 3 (4)2 2(4) 3 16 8 3 8 3 5 Substitute 4 for x. When x 4, x2 2x 3 is positive. Therefore, x2 2x 3 will be positive for all values of x in this interval. Indicate this on the number line. Look at the number line above. The solution set of x2 2x 3 0 consists of the interval(s) where x2 2x 3 is negative. This is in Interval B, (1, 3). The graph of the solution set is 5 2 1 1 2 5 The solution set is (1, 3). This is the same as the result we obtained in Example 1 by graphing. Read this example very carefully. YOU TRY 2 Solve x2 5x 4 0. Graph the solution set, and write the solution in interval notation. Next we will summarize how to solve a quadratic inequality. www.mhhe.com/messersmith SECTION 10.7 Quadratic and Rational Inequalities 683


messersmith_power_intermediate_algebra_1e_ch4_7_10
To see the actual publication please follow the link above