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messersmith_power_introductory_algebra_1e_ch4_7_10

Note Since both terms in 9v2 54v are divisible by 9, we could have started part b) by dividing by 9: 9v 2 9 54v 9 Divide by 9. v 2 6v v 2 6v 0 Write in standard form. v (v 6) 0 Factor. b R v 0  or  v 6 0 Set each factor equal to zero. d 6 Solve. The solution set is {6, 0}. We get the same result. We cannot divide by v even though each term contains a factor of v. Doing so would eliminate the solution of zero. In general, we can divide an equation by a nonzero real number but we cannot divide an equation by a variable because we may eliminate a solution, and we may be dividing by zero. c) To solve h2 5(2h 5), begin by writing the equation in standard form. h2 10h 25 Distribute. h2 10h 25 0 Write in standard form. (h 5)2 0 Factor. Since (h 5)2 0 means (h 5)(h 5) 0, setting each factor equal to zero will result in the same value for h. h 5 0 Set h 5 0. h 5 Solve. Check. The solution set is {5}. d) It is tempting to solve (w 4)(w 5) 2 like this: (w 4)(w 5) 2 b R w 4 2 or w 5 2 This is incorrect! One side of the equation must equal zero in order to use the zero product rule. Begin by multiplying on the left. (w 4)(w 5) 2 w2 9w 20 2 Multiply using FOIL. w2 9w 18 0 Standard form (w 6)(w 3) 0 Factor. b R w 6 0 or w 3 0 Set each factor equal to zero. w 6 or w 3 Solve. The check is left to the student. The solution set is {3, 6}. Complete all the You Try problems. Ask for help if you get stuck. www.mhhe.com/messersmith SECTION 7.5 Solving Quadratic Equations by Factoring 427


messersmith_power_introductory_algebra_1e_ch4_7_10
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