Algebra 2 Final Exam Review: Equations and Expressions Solve

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) βˆ’ 2(π‘₯ βˆ’ 3) = π‘₯ + 6
8π‘₯ + 24 βˆ’ 2π‘₯ + 6 = π‘₯ + 6
6π‘₯ + 30 = π‘₯ + 6
5π‘₯ = βˆ’24
24
π‘₯=βˆ’5
2.
2
π‘₯ = 40
5
5 2
5
βˆ™ π‘₯ = 2 βˆ™ 40
2 5
π‘₯ = 100
3.
5
6
3
π‘₯+7=4
5
3
12 (6 π‘₯ + 7 = 4)
10π‘₯ + 84 = 9
10π‘₯ = βˆ’75
π‘₯ = βˆ’7.5
4. Solve by factoring.
π‘₯ 2 + 11π‘₯ + 14 = 74
π‘₯ 2 + 11π‘₯ βˆ’ 60 = 0
(π‘₯ + 15)(π‘₯ βˆ’ 4) = 0
π‘₯ = βˆ’15, π‘₯ = 4
5. Solve by factoring.
8π‘₯ 2 = 20π‘₯
8π‘₯ 2 βˆ’ 20π‘₯ = 0
4π‘₯(2π‘₯ βˆ’ 5) = 0
5
π‘₯ = 0, π‘₯ = 2
6. Solve by completing the square.
π‘₯ 2 βˆ’ 10π‘₯ + 85 = 11
π‘₯ 2 βˆ’ 10π‘₯ + ______ = βˆ’74 + _______
π‘₯ 2 βˆ’ 10π‘₯ + 25 = βˆ’74 + 25
(π‘₯ βˆ’ 5)2 = βˆ’49
π‘₯ βˆ’ 5 = ±7𝑖
π‘₯ = 5 ± 7𝑖
7. Solve using the quadratic formula
3π‘₯ 2 + 5π‘₯ + 8 = 7
3π‘₯ 2 + 5π‘₯ + 1 = 0
π‘₯=
βˆ’5±βˆš52 βˆ’4(3)(1)
2(3)
βˆ’5±βˆš13
π‘₯= 6
π‘₯ β‰ˆ βˆ’0.23, π‘₯ β‰ˆ βˆ’1.43
8. π‘₯ 3 + 5π‘₯ 2 βˆ’ 16π‘₯ = 80
π‘₯ 3 + 5π‘₯ 2 βˆ’ 16π‘₯ βˆ’ 80 = 0
π‘₯ 2 (π‘₯ + 5) βˆ’ 16(π‘₯ + 5) = 0
(π‘₯ + 5)(π‘₯ 2 βˆ’ 16) = 0
(π‘₯ + 5)(π‘₯ βˆ’ 4)(π‘₯ + 4) = 0
π‘₯ = βˆ’5, π‘₯ = 4, π‘₯ = βˆ’4
9. π‘₯ 4 βˆ’ π‘₯ 2 = 90
π‘₯ 4 βˆ’ π‘₯ 2 βˆ’ 90 = 0
(π‘₯ 2 + 9)(π‘₯ 2 βˆ’ 10) = 0
π‘₯ 2 + 9 = 0, π‘₯ 2 βˆ’ 10 = 0
π‘₯ 2 = βˆ’9, π‘₯ 2 = 10
π‘₯ = ±3𝑖, π‘₯ = ±βˆš10 β‰ˆ ±3.16
10. 2π‘₯ 4 + 130π‘₯ = 2π‘₯
2π‘₯ 4 + 128π‘₯ = 0
2π‘₯(π‘₯ 3 + 64) = 0
2π‘₯(π‘₯ + 4)(π‘₯ 2 βˆ’ 4π‘₯ + 16) = 0
π‘₯ = 0, π‘₯ = βˆ’4, π‘₯ = 2 ± 2√3𝑖
11. |2π‘₯ + 3| = 17
2π‘₯ + 3 = βˆ’17
2π‘₯ = βˆ’20
π‘₯ = βˆ’10
2π‘₯ + 3 = 17
2π‘₯ = 14
π‘₯=7
12. |6 βˆ’ 5π‘₯| = 21
6 βˆ’ 5π‘₯ = βˆ’21
βˆ’5π‘₯ = βˆ’27
π‘₯ = 5.4
6 βˆ’ 5π‘₯ = 21
βˆ’5π‘₯ = 15
π‘₯ = βˆ’3
13. 200(1.13)π‘₯ = 1300
1.13π‘₯ = 6.5
π‘₯ = log1.13 6.5
π‘₯ β‰ˆ 15.315
π‘₯ 2 βˆ’ 4π‘₯ + 16 = 0
π‘₯ 2 βˆ’ 4π‘₯ + 4 = βˆ’16 + 4
(π‘₯ βˆ’ 2)2 = βˆ’12
π‘₯ βˆ’ 2 = ±βˆšβˆ’12
π‘₯ βˆ’ 2 = ±2√3𝑖
14. 17500(0.85)π‘₯ = 7000
0.85π‘₯ = 0.4
π‘₯ = log 0.85 0.4
π‘₯ β‰ˆ 5.638
Solve each system of equations using the method of your choice.
15. 𝑦 = 7π‘₯ + 10
6π‘₯ + 2𝑦 = 96
6π‘₯ + 2(7π‘₯ + 10) = 96
6π‘₯ + 14π‘₯ + 20 = 96
20π‘₯ = 76
π‘₯ = 3.8
𝑦 = 7π‘₯ + 10
𝑦 = 7(3.8) + 10
𝑦 = 36.6
(3.8, 36.6)
π‘₯ + 4(1) = 17
π‘₯ + 4 = 17
π‘₯ = 13
(13, 1)
16. 5π‘₯ βˆ’ 3𝑦 = 62
π‘₯ + 4𝑦 = 17
5π‘₯ βˆ’ 3𝑦 = 62
βˆ’5π‘₯ βˆ’ 20𝑦 = βˆ’85
βˆ’23𝑦 = βˆ’23
𝑦=1
17. π‘₯ + 𝑦 βˆ’ 𝑧 = βˆ’9
π‘₯βˆ’π‘¦+𝑧 =1
βˆ’π‘₯ + 𝑦 + 𝑧 = 11
π‘₯
βˆ’4
1
1 βˆ’1 βˆ’1 βˆ’9
[𝑦] = [ 1 βˆ’1 1 ] βˆ™ [ 1 ] = [ 1 ]
𝑧
βˆ’1 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
βˆ’ π‘₯ 2 +2π‘₯
π‘₯ 2 βˆ’4
βˆ’ π‘₯(π‘₯+2) = π‘₯(π‘₯+2) βˆ’ π‘₯(π‘₯+2) = π‘₯(π‘₯+2) =
π‘₯+2
(π‘₯+2)(π‘₯βˆ’2)
π‘₯(π‘₯+2)
=
π‘₯βˆ’2
π‘₯
(βˆ’4,1,6)
21. Simplify completely:
6π‘₯+4
π‘₯βˆ’3
2
3+
π‘₯
π‘₯(π‘₯βˆ’3)
6π‘₯+4
π‘₯βˆ’3
2
3+
π‘₯
6π‘₯ 2 +4π‘₯
2π‘₯(3π‘₯+2)
2π‘₯(3π‘₯+2)
2π‘₯
βˆ™ π‘₯(π‘₯βˆ’3) = 3π‘₯ 2 βˆ’9π‘₯+2π‘₯βˆ’6 = 3π‘₯ 2 βˆ’7π‘₯βˆ’6 = (3π‘₯+2)(π‘₯βˆ’3) = π‘₯βˆ’3
Factor completely.
22. 4π‘₯ 5 βˆ’ 500π‘₯ 2
4π‘₯ 2 (π‘₯ 3 βˆ’ 125)
4π‘₯ 2 (π‘₯ βˆ’ 5)(π‘₯ 2 + 5π‘₯ + 25)
23. π‘₯ 4 βˆ’ 37π‘₯ 2 + 36
(π‘₯ 2 βˆ’ 36)(π‘₯ 2 βˆ’ 1)
(π‘₯ + 6)(π‘₯ βˆ’ 6)(π‘₯ + 1)(π‘₯ βˆ’ 1)
24. 15π‘₯ 3 βˆ’ 114π‘₯ 2 βˆ’ 360π‘₯
3π‘₯(5π‘₯ 2 βˆ’ 38π‘₯ βˆ’ 120)
3π‘₯(5π‘₯ + 12)(π‘₯ βˆ’ 10)
25. π‘₯ 5 βˆ’ 100π‘₯ 3 + 1000π‘₯ 2 βˆ’ 100000
π‘₯ 3 (π‘₯ 2 βˆ’ 100) + 1000(π‘₯ 2 βˆ’ 100)
(π‘₯ 2 βˆ’ 100)(π‘₯ 3 + 1000)
(π‘₯ + 10)(π‘₯ βˆ’ 10)(π‘₯ + 10)(π‘₯ 2 βˆ’ 10π‘₯ + 100)