Algebra 2 Final Exam Review: Equations and Expressions Solve each equation. Keep rational solutions in exact form, and round off irrational solutions to the nearest hundredth. 1. 8(π₯ + 3) β 2(π₯ β 3) = π₯ + 6 8π₯ + 24 β 2π₯ + 6 = π₯ + 6 6π₯ + 30 = π₯ + 6 5π₯ = β24 24 π₯=β5 2. 2 π₯ = 40 5 5 2 5 β π₯ = 2 β 40 2 5 π₯ = 100 3. 5 6 3 π₯+7=4 5 3 12 (6 π₯ + 7 = 4) 10π₯ + 84 = 9 10π₯ = β75 π₯ = β7.5 4. Solve by factoring. π₯ 2 + 11π₯ + 14 = 74 π₯ 2 + 11π₯ β 60 = 0 (π₯ + 15)(π₯ β 4) = 0 π₯ = β15, π₯ = 4 5. Solve by factoring. 8π₯ 2 = 20π₯ 8π₯ 2 β 20π₯ = 0 4π₯(2π₯ β 5) = 0 5 π₯ = 0, π₯ = 2 6. Solve by completing the square. π₯ 2 β 10π₯ + 85 = 11 π₯ 2 β 10π₯ + ______ = β74 + _______ π₯ 2 β 10π₯ + 25 = β74 + 25 (π₯ β 5)2 = β49 π₯ β 5 = ±7π π₯ = 5 ± 7π 7. Solve using the quadratic formula 3π₯ 2 + 5π₯ + 8 = 7 3π₯ 2 + 5π₯ + 1 = 0 π₯= β5±β52 β4(3)(1) 2(3) β5±β13 π₯= 6 π₯ β β0.23, π₯ β β1.43 8. π₯ 3 + 5π₯ 2 β 16π₯ = 80 π₯ 3 + 5π₯ 2 β 16π₯ β 80 = 0 π₯ 2 (π₯ + 5) β 16(π₯ + 5) = 0 (π₯ + 5)(π₯ 2 β 16) = 0 (π₯ + 5)(π₯ β 4)(π₯ + 4) = 0 π₯ = β5, π₯ = 4, π₯ = β4 9. π₯ 4 β π₯ 2 = 90 π₯ 4 β π₯ 2 β 90 = 0 (π₯ 2 + 9)(π₯ 2 β 10) = 0 π₯ 2 + 9 = 0, π₯ 2 β 10 = 0 π₯ 2 = β9, π₯ 2 = 10 π₯ = ±3π, π₯ = ±β10 β ±3.16 10. 2π₯ 4 + 130π₯ = 2π₯ 2π₯ 4 + 128π₯ = 0 2π₯(π₯ 3 + 64) = 0 2π₯(π₯ + 4)(π₯ 2 β 4π₯ + 16) = 0 π₯ = 0, π₯ = β4, π₯ = 2 ± 2β3π 11. |2π₯ + 3| = 17 2π₯ + 3 = β17 2π₯ = β20 π₯ = β10 2π₯ + 3 = 17 2π₯ = 14 π₯=7 12. |6 β 5π₯| = 21 6 β 5π₯ = β21 β5π₯ = β27 π₯ = 5.4 6 β 5π₯ = 21 β5π₯ = 15 π₯ = β3 13. 200(1.13)π₯ = 1300 1.13π₯ = 6.5 π₯ = log1.13 6.5 π₯ β 15.315 π₯ 2 β 4π₯ + 16 = 0 π₯ 2 β 4π₯ + 4 = β16 + 4 (π₯ β 2)2 = β12 π₯ β 2 = ±ββ12 π₯ β 2 = ±2β3π 14. 17500(0.85)π₯ = 7000 0.85π₯ = 0.4 π₯ = log 0.85 0.4 π₯ β 5.638 Solve each system of equations using the method of your choice. 15. π¦ = 7π₯ + 10 6π₯ + 2π¦ = 96 6π₯ + 2(7π₯ + 10) = 96 6π₯ + 14π₯ + 20 = 96 20π₯ = 76 π₯ = 3.8 π¦ = 7π₯ + 10 π¦ = 7(3.8) + 10 π¦ = 36.6 (3.8, 36.6) π₯ + 4(1) = 17 π₯ + 4 = 17 π₯ = 13 (13, 1) 16. 5π₯ β 3π¦ = 62 π₯ + 4π¦ = 17 5π₯ β 3π¦ = 62 β5π₯ β 20π¦ = β85 β23π¦ = β23 π¦=1 17. π₯ + π¦ β π§ = β9 π₯βπ¦+π§ =1 βπ₯ + π¦ + π§ = 11 π₯ β4 1 1 β1 β1 β9 [π¦] = [ 1 β1 1 ] β [ 1 ] = [ 1 ] π§ β1 1 1 11 6 (π₯ + 4)3 18. Expand the power: (π₯ + 4)(π₯ + 4)(π₯ + 4) (π₯ + 4)(π₯ 2 + 8π₯ + 16) π₯ 3 + 8π₯ 2 + 16π₯ + 4π₯ 2 + 32π₯ + 64 π₯ 3 + 12π₯ 2 + 48π₯ + 64 (π₯ 2 + 3π₯ + 5)2 19. Expand the power: (π₯ 2 + 3π₯ + 5)(π₯ 2 + 3π₯ + 5) π₯ 4 + 3π₯ 3 + 5π₯ 2 + 3π₯ 3 + 9π₯ 2 + 15π₯ + 5π₯ 2 + 15π₯ + 25 π₯ 4 + 6π₯ 3 + 19π₯ 2 + 30π₯ + 25 20. Simplify completely: π₯ 4 π₯2 π₯ π₯+2 4 4 β π₯ 2 +2π₯ π₯ 2 β4 β π₯(π₯+2) = π₯(π₯+2) β π₯(π₯+2) = π₯(π₯+2) = π₯+2 (π₯+2)(π₯β2) π₯(π₯+2) = π₯β2 π₯ (β4,1,6) 21. Simplify completely: 6π₯+4 π₯β3 2 3+ π₯ π₯(π₯β3) 6π₯+4 π₯β3 2 3+ π₯ 6π₯ 2 +4π₯ 2π₯(3π₯+2) 2π₯(3π₯+2) 2π₯ β π₯(π₯β3) = 3π₯ 2 β9π₯+2π₯β6 = 3π₯ 2 β7π₯β6 = (3π₯+2)(π₯β3) = π₯β3 Factor completely. 22. 4π₯ 5 β 500π₯ 2 4π₯ 2 (π₯ 3 β 125) 4π₯ 2 (π₯ β 5)(π₯ 2 + 5π₯ + 25) 23. π₯ 4 β 37π₯ 2 + 36 (π₯ 2 β 36)(π₯ 2 β 1) (π₯ + 6)(π₯ β 6)(π₯ + 1)(π₯ β 1) 24. 15π₯ 3 β 114π₯ 2 β 360π₯ 3π₯(5π₯ 2 β 38π₯ β 120) 3π₯(5π₯ + 12)(π₯ β 10) 25. π₯ 5 β 100π₯ 3 + 1000π₯ 2 β 100000 π₯ 3 (π₯ 2 β 100) + 1000(π₯ 2 β 100) (π₯ 2 β 100)(π₯ 3 + 1000) (π₯ + 10)(π₯ β 10)(π₯ + 10)(π₯ 2 β 10π₯ + 100)
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