YOU TRY 2 Solve (t 3)2 4 using the square root property. Solve. a) (w 4)2 3 b) (3r 2)2 25 c) (2x 7)2 18 Solution a) (w 4)2 3 w 4 13 Square root property w 4 13 Add 4 to each side. Check: w 4 13: (w 4)2 3 w 4 13: (w 4)2 3 (4 13 4)2 3 (4 13 4)2 3 (13)2 3 (13)2 3 3 3 ✓ 3 3 ✓ The solution set is {4 13, 4 13} or {4 13}. b) (3r 2)2 25 3r 2 125 Square root property 3r 2 5 This means 3r 2 5 or 3r 2 5. Solve both equations. 3r 2 5 or 3r 2 5 3r 3 3r 7 Subtract 2 from each side. r 1 or r 7 3 Divide by 3. The check is left to the student. The solution set is e 7 3 , 1 f . c) (2x 7)2 18 2x 7 118 Square root property 2x 7 312 Simplify 118. 2x 7 312 Add 7 to each side. 7 312 x 2 Divide by 2. One solution is 7 312 2 , and the other is 7 312 2 . The solution set is e 7 312 2 , 7 312 2 f . This can also be written as e 7 312 2 f . The check is left to the student. EXAMPLE 3 In-Class Example 3 Solve. a) (k 6)2 5 b) (2f 5)2 16 c) (4d 7)2 80 Answer: a) {6 15, 6 15} b) e 9 2 , 1 2 f c) e 7 415 4 , 7 415 4 f www.mhhe.com/messersmith SECTION 10.1 Solving Quadratic Equations Using the Square Root Property 609
messersmith_power_introductory_algebra_1e_ch4_7_10
To see the actual publication please follow the link above