Section 2.6 Absolute Value Equations 121 3b. Find the absolute error when the value of 3.14 is used for π. Round the answer to four decimal places. Solution 3b. We can find the absolute error using the given formula since we know the approximated value is 3.14 and the exact value is π. So, a = 3.14 and x = π. Eabs = ux - au Eabs = uπ - 3.14u Eabs = 0.0016 So, the absolute error is 0.0016. Student Check 3 Solve each problem. a. Suppose a dog measures 80 cm long with a 3 cm absolute error. Find the possible values for the exact length of the dog. b. Find the absolute error when the value of 2.7 is used for the number e, where e = 2.71828 . . . . Round the answer to four decimal places. Troubleshooting Common Errors Some common errors associated with solving absolute value equations are shown. Objective 4 Examples A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 4a. Solve: u2x - 3u = 7. Incorrect Solution Correct Solution and Explanation There are two solutions of this equation since there are two numbers whose u2x - 3u = 7 absolute value is 7. 2x - 3 = 7 2x = 10 u2x - 3u = 7 x = 5 2x - 3 = 7 or 2x - 3=-7 2x =10 2x=-4 The solution set is 556. x =5 x=-2 The solution set is 5-2, 56. 4b. Solve: uy + 2u - 1 = 4. Incorrect Solution Correct Solution and Explanation The absolute value expression must be isolated prior to solving. We must first add 1 each side. uy + 2u - 1 = 4 uy + 2u = 5 y + 2 = 5 or y + 2=-5 y = 3 y=-7 The solution set is 5-7, 36. uy + 2u -1 = 4 y + 2 - 1 = 4 or y + 2 - 1=-4 y 1 = 4 y 1=-4 =3 y=-5 The solution set is 5-5, 36. Objective 4 ▶ Troubleshoot common errors. u2 on s 4 + + y t is 5 5
hendricks_intermediate_algebra_1e_ch1_3
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