College Algebra Fall 2015 Final Exam Prep – Solutions

1. Simplify 3√−8

= 3√−1 ∙ √8

= 3𝑖√4 ∙ √2

= 3𝑖 ∙ 2√2

= 6𝑖√2

1. 6𝑖√2

2. Simplify 𝑖85

85

4= 21

1

4; note the remainder is 1

𝑖85 = 𝑖1

2. 𝑖

3. Simplify 19 − (3 − 14𝑖) + 9𝑖

= 19 − 3 + 14𝑖 + 9𝑖

= 16 + 23𝑖

3. 16 + 23𝑖

4. Simplify (4 − 3𝑖)(4 + 3𝑖)

= 42 + 32

= 25

4. 25

5. Simplify (1 − 2𝑖)3

= [(1 − 2𝑖)(1 − 2𝑖)](1 − 2𝑖)

= (1 − 4𝑖 + 4𝑖2)(1 − 2𝑖)

= (1 − 4𝑖 + 4(−1))(1 − 2𝑖)

= (−3 − 4𝑖)(1 − 2𝑖)

= −3 + 2𝑖 + 8𝑖2

= −3 + 2𝑖 + 8(−1)

= −11 + 2𝑖

5. −11 + 2𝑖

6. Simplify 6+21𝑖

3𝑖

= (6 + 21𝑖

0 + 3𝑖) (

0 − 3𝑖

0 − 3𝑖)

=−18𝑖 − 63𝑖2

−9𝑖2

=−18𝑖 − 63(−1)

−9(−1)

=63 − 18𝑖

9

=63

9−

18

9𝑖

= 7 − 2𝑖

6. 7 − 2𝑖

7. Simplify 3−5𝑖

8−36𝑖

= (3 − 5𝑖

8 − 36𝑖) (

8 + 36𝑖

8 + 36𝑖)

=24 + 68𝑖 − 180𝑖2

82 + 362

=24 + 68𝑖 − 180(−1)

1360

=204 + 68𝑖

1360

=204

1360+

68

1360𝑖

=3 ∙ 68

20 ∙ 68+

1 ∙ 68

20 ∙ 68𝑖

=3

20+

1

20𝑖

7. 3

20+

1

20𝑖

8. Simplify 23−11𝑖

5+𝑖−

3+37𝑖

5+𝑖

=23 − 11𝑖 − 3 − 37𝑖

5 + 𝑖

=20 − 48𝑖

5 + 𝑖

= (20 − 48𝑖

5 + 1𝑖) (

5 − 1𝑖

5 − 1𝑖)

=100 − 260𝑖 + 48𝑖2

52 + 12

=100 − 260𝑖 + 48(−1)

26

=52 − 260𝑖

26

=52

26−

260

26𝑖

= 2 − 10𝑖

8. 2 − 10𝑖

9. Plot −5 + 2𝑖

10. Find |11 + 3𝑖|

= √112 + 32

= √130

10. √130

11. Write the complex number that is plotted below.

11. −5 + 2𝑖

12. Mixture - Amy wants to make 14 gallons of a 54% alcohol

solution by mixing together a 63% alcohol solution and pure water. How many gallons of pure water must she use?

. 63(14 − 𝑥) = .54(14) . 63(14) − .63𝑥 = .54(14) 8.82 − .63𝑥 = 7.56 −.63𝑥 = −1.26 𝑥 = 2 She must use 2 gallons of pure water.

12. ________________________________

13. Uniform Motion - Brenda left the hospital and drove south.

Mandy left two hours later driving 16 mph faster in an effort to catch up to her. After three hours Mandy finally caught up. What was Brenda’s average speed? Let 𝑥 = 𝑅Brenda 𝐷Brenda = 𝐷Mandy

𝑅Brenda ∙ 𝑇Brenda = 𝑅Mandy ∙ 𝑇Mandy

𝑥(5) = (𝑥 + 16)(3)

5𝑥 = 3𝑥 + 48

2𝑥 = 48

𝑥 = 24 mph (Brenda)

13. 𝑥 = 24 mph (Brenda)

14. Work - Working alone, Ted can sweep a porch in 11 minutes. One day his friend Cindy helped him and it only took 5.74 minutes. Find how long it would take Cindy to do it alone. (Round to the nearest hundredth.) Let 𝑥 = length of time for Cindy to sweep the porch alone 1

11+

1

𝑥=

1

5.74

(1

11) (

5.74𝑥

5.74𝑥) + (

1

𝑥) (

11(5.74)

11(5.74)) = (

1

5.74) (

11𝑥

11𝑥)

5.74𝑥 + 11(5.74) = 11𝑥

5.74𝑥 + 63.14 = 11𝑥

63.14 = 5.26𝑥

𝑥 = 12

14. 𝑥 = 12

15. Solve 4𝑥2 + 9 = −5

(Complex plane) 4𝑥2 = −14

𝑥2 =−14

4

𝑥 = ±√−14

4

𝑥 =±𝑖√14

2

15. 𝑥 =

±𝑖√14

2

16. Solve 𝑥2 = −3𝑥 + 4

(Complex plane) 𝑥2 + 3𝑥 − 4 = 0

(𝑥 + 4)(𝑥 − 1) = 0

𝑥 = −4, 𝑥 = 1

16. 𝑥 = −4, 𝑥 = 1

17. Solve 5𝑥2 = −5 + 𝑥 (Complex Plane) 5𝑥2 − 𝑥 + 5 = 0

𝑎 = 5, 𝑏 = −1, 𝑐 = 5

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑥 =−(−1) ± √(−1)2 − 4(5)(5)

2(5)

𝑥 =1 ± √−99

10

𝑥 =1 ± √−1 ∙ √9 ∙ √11

10

𝑥 =1 ± 3𝑖√11

10

𝑥 =1

10±

3√11

10𝑖

17. ________________________________

18. Solve by completing the square 𝑥2 − 4𝑥 − 42 = 8.

(Complex plane) (You must clearly show each step) 𝑥2 − 4𝑥 = 50

𝑥2 − 4𝑥 + 4 = 50 + 4

(𝑥 − 2)2 = 54

𝑥 − 2 = ±√54

𝑥 = 2 ± √9 ∙ √6

𝑥 = 2 ± 3√6

17. 𝑥 = 2 ± 3√6

19. Find the discriminant and state the number and type of solution. 10𝑥2 + 5𝑥 − 11 = −3 (Complex plane) 10𝑥2 + 5𝑥 − 8 = 0

𝑎 = 10, 𝑏 = 5, 𝑐 = −8

discriminant = 𝑏2 − 4𝑎𝑐

discriminant = 52 − 4(10)(−8)

discriminant = 345

Since the discriminant is positive and not a perfect square,

there are two real irrational solutions

19. discriminant = 345

Since the discriminant is positive and not a perfect square, there are two real irrational solutions

20. Solve

1

𝑥+

1

𝑥2+5𝑥=

4

𝑥2+5𝑥

(Must check for extraneous solutions) 1

𝑥+

1

𝑥(𝑥 + 5)=

4

𝑥(𝑥 + 5)

(1

𝑥) (

𝑥 + 5

𝑥 + 5) +

1

𝑥(𝑥 + 5)=

4

𝑥(𝑥 + 5)

𝑥 + 5 + 1 = 4

𝑥 = −2

Check:

1

−2+

1

−2(−2 + 5) ?

4

−2(−2 + 5)

−1

2+

1

−6 ?

4

−6

−3

6−

1

6 ? −

4

6

−4

6= −

4

6; it checks

𝑥 = −2

20. ________________________________

21. Solve 𝑥−5

𝑥2−3𝑥= 1 −

𝑥+2

𝑥

(Must check for extraneous solutions)

𝑥 − 5

𝑥2 − 3𝑥= 1 −

𝑥 + 2

𝑥

𝑥 − 5

𝑥(𝑥 − 3)=

𝑥(𝑥 − 3)

𝑥(𝑥 − 3)−

(𝑥 + 2)(𝑥 − 3)

𝑥(𝑥 − 3)

𝑥 − 5 = 𝑥2 − 3𝑥 − (𝑥2 − 𝑥 − 6)

𝑥 − 5 = −3𝑥 + 𝑥 + 6

−11 = −3𝑥

𝑥 =11

3

Check:

113 − 5

(113 )

2

− 3 (113 )

? 1 −

113 + 2

113

−6

11=

−6

11

𝑥 =11

3

21. 𝑥 =

11

3

22. Solve

5𝑥+1

𝑥+3−

𝑥+4

𝑥+3= 0

(Must check for extraneous solutions)

5𝑥 + 1 − (𝑥 + 4)

𝑥 + 3= 0

4𝑥 − 3

𝑥 + 3= 0

4𝑥 − 3 = 0

𝑥 =3

4

Check

5 (34) + 1

34 + 3

−

34 + 4

34 + 3

? 0

0 = 0

𝑥 =3

4

22. 𝑥 =

3

4

23. Solve 𝑥 − 8 = √43 − 6𝑥 (Must check for extraneous solutions)

(𝑥 − 8)2 = (√43 − 6𝑥)2

𝑥2 − 16𝑥 + 64 = 43 − 6𝑥

𝑥2 − 10𝑥 + 21 = 0

(𝑥 − 3)(𝑥 − 7) = 0

𝑥 = 3, 𝑥 = 7

Check

3 − 8 ? √43 − 6(3)

−5 ≠ √25

𝑥 = 3 is an extraneous solution

7 − 8 ? √43 − 6(7)

−1 ≠ √1

𝑥 = 7 is an extraneous solution

No Solution

23. No Solution

24. Solve √8 − 𝑥 − √1 − 3𝑥 = −1 (Must check for extraneous solutions)

√8 − 𝑥 = √1 − 3𝑥 − 1

(√8 − 𝑥)2

= (√1 − 3𝑥 − 1)2

8 − 𝑥 = 1 − 3𝑥 − 2√1 − 3𝑥 + 1

8 − 𝑥 = 2 − 3𝑥 − 2√1 − 3𝑥

6 + 2𝑥 = 2√1 − 3𝑥

3 + 𝑥 = √1 − 3𝑥

(3 + 𝑥)2 = (√1 − 3𝑥)2

9 + 6𝑥 + 𝑥2 = 1 − 3𝑥

𝑥2 + 9𝑥 + 8 = 0

(𝑥 + 8)(𝑥 + 1) = 0

𝑥 = −8, 𝑥 = −1

Check

√8 − (−8) − √1 − 3(−8) ? −1

√16 − √25 ? −1

4 − 5 = −1

√8 − (−1) − √1 − 3(−1) ? −1

√9 − √4 ? −1

3 − 2 ≠ −1

𝑥 = −1 is an extraneous solution

𝑥 = −8

24. 𝑥 = −8

25. Solve √2𝑥 − 5 = −2 + √4𝑥 − 3 (Must check for extraneous solutions)

(√2𝑥 − 5)2

= (−2 + √4𝑥 − 3)2

2𝑥 − 5 = 4 − 4√4𝑥 − 3 + 4𝑥 − 3

2𝑥 − 5 = 1 + 4𝑥 − 4√4𝑥 − 3

−2𝑥 − 6 = −4√4𝑥 − 3

𝑥 + 3 = 2√4𝑥 − 3

(𝑥 + 3)2 = (2√4𝑥 − 3)2

𝑥2 + 6𝑥 + 9 = 4(4𝑥 − 3)

𝑥2 + 6𝑥 + 9 = 16𝑥 − 12

𝑥2 − 10𝑥 + 21 = 0

(𝑥 − 3)(𝑥 − 7) = 0

𝑥 = 3, 𝑥 = 7

Check

√2(3) − 5 ? −2 + √4(3) − 3

√6 − 5 ? −2 + √12 − 3

1 ? −2 + 3

1 = 1

√2(7) − 5 ? −2 + √4(7) − 3

√9 ? −2 + √25

3 = 3

𝑥 = 3, 𝑥 = 7

25. 𝑥 = 3, 𝑥 = 7

26. Solve {

−6𝑥 + 2𝑦 − 6𝑧 = −12 −6𝑥 + 3𝑦 − 𝑧 = 14

−5𝑥 − 2𝑦 − 𝑧 = −16

Equation 1Equation 2Equation 3

Eliminate z Eqn 1 − 6 ∙ (Eqn 2) −6𝑥 + 2𝑦 − 6𝑧 = −12 36𝑥 − 18𝑦 + 6𝑧 = −84 30𝑥 − 16𝑦 = −96 15𝑥 − 8𝑦 = −48 Equation 4 Eliminate z Eqn 1 − 6 ∙ (Eqn 3) −6𝑥 + 2𝑦 − 6𝑧 = −12 30𝑥 + 12𝑦 + 6𝑧 = 96 24𝑥 + 14𝑦 = 84 12𝑥 + 7𝑦 = 42 Equation 5 Eliminate y 7 ∙ (Eqn 4) + 8 ∙ (Eqn 5) 105𝑥 − 56𝑦 = −336 96𝑥 + 56𝑦 = 336 201𝑥 = 0 𝑥 = 0 15(0) − 8𝑦 = −48 𝑦 = 6 −6(0) + 2(6) − 6𝑧 = −12 −6𝑧 = −24 𝑧 = 4 (𝑥, 𝑦, 𝑧) = (0, 6, 4)

26. (𝑥, 𝑦, 𝑧) = (0, 6, 4)

28. Determine the domain and range of the relation 𝐴 = {(1, 3), (3, 4), (5, 3), (3, 6)}

28. 𝐷 = {1, 3, 5} 𝑅 = {3, 4, 6}

29. Determine the domain and range of the relation.

(Express your answer using interval notation)

29. 𝐷 = [1, ∞) 𝑅 = [2, ∞)

30. Is the relation {(1, 3), (−3, 4), (5, 3), (3, 6)} a function? 30. Yes

31. Is the relation a function?

31. No

32. Does the equation |𝑥2 − 4| = 5𝑦 define y as a function of x? 32. Yes

33. Does the equation 𝑥 ± 𝑦 = 9 define y as a function of x? 33. No

34. Does the equation 15𝑥 = 𝑦2 + 1 define y as a function of x? 34. No

35. Find the domain of the function, 𝑓(𝑥) = 𝑥2 + 9𝑥 + 20 35. 𝐷 = (−∞, ∞)

36. Find the domain of the function 𝑓(𝑥) = √𝑥 − 10 + 3

𝑥 − 10 ≥ 0

𝑥 ≥ 10

𝐷 = [10, ∞)

36. 𝐷 = [10, ∞)

37. Find the domain of the function 𝑓(𝑥) =𝑥+1

𝑥2−9

𝑥2 − 9 ≠ 0

(𝑥 + 3)(𝑥 − 3) = 0

𝑥 = −3, 𝑥 = 3

𝐷 = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)

37. 𝐷 = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)

38. Find the domain of the function 𝑓(𝑥) =

√𝑥+5

𝑥2+5𝑥+6

𝑥 + 5 ≥ 0 and 𝑥2 + 5𝑥 + 6 ≠ 0

𝑥 ≥ −5 (𝑥 + 3)(𝑥 + 2) = 0

𝑥 = −3, 𝑥 = −2

𝐷 = [−5, −3) ∪ (−3, −2) ∪ (−2, ∞)

38. 𝐷 = [−5, −3) ∪ (−3, −2) ∪ (−2, ∞)

39. Find the domain of the function 𝑓(𝑥) =

2𝑥

√1−𝑥

1 − 𝑥 > 0

1 > 𝑥

𝐷 = (−∞, 1)

39. 𝐷 = (−∞, 1)

42. Let 𝑓(𝑥) =

𝑥+2

𝑥−5 and 𝑔(𝑥) =

𝑥−1

𝑥+4, find the domain of

(𝑓𝑔)(𝑥) 𝐷𝑓: 𝑥 ≠ 5

𝐷𝑔: 𝑥 ≠ −4

𝐷𝑓𝑔 = 𝐷𝑓 ∩ 𝐷𝑔

𝐷𝑓𝑔 = (−∞, −4) ∪ (−4, 5) ∪ (5, ∞)

42. 𝐷𝑓𝑔 = (−∞, −4) ∪ (−4, 5) ∪ (5, ∞)

43. Let 𝑓(𝑥) =

𝑥+2

𝑥−5 and 𝑔(𝑥) =

𝑥−1

𝑥+4,

find the domain of (𝑓

𝑔) (𝑥)

𝐷𝑓: 𝑥 ≠ 5

𝐷𝑔: 𝑥 ≠ −4

𝑔(𝑥) ≠ 0 ⇒ 𝑥 ≠ 1

𝐷𝑓𝑔

= (−∞, −4) ∪ (−4, 1) ∪ (1, 5) ∪ (5, ∞)

43. 𝐷𝑓𝑔

= (−∞, −4) ∪ (−4, 1) ∪ (1, 5) ∪ (5, ∞)

44. Let 𝑓(𝑥) =𝑥+2

𝑥−5 and 𝑔(𝑥) =

𝑥−1

𝑥+4,

find the domain of (𝑔

𝑓) (𝑥)

𝐷𝑓: 𝑥 ≠ 5

𝐷𝑔: 𝑥 ≠ −4

𝑓(𝑥) ≠ 0, ⇒ 𝑥 ≠ −2

𝐷𝑔𝑓

= (−∞, −4) ∪ (−4, −2) ∪ (−2, 5) ∪ (5, ∞)

44. 𝐷𝑔𝑓

= (−∞, −4) ∪ (−4, −2) ∪ (−2, 5) ∪ (5, ∞)

45. Let 𝑓(𝑥) =

𝑥+2

𝑥−5 and 𝑔(𝑥) =

𝑥−1

𝑥+4, find the

domain of (𝑓 ∘ 𝑔)(𝑥) 𝑥 ∈ 𝐷𝑔 and 𝑔(𝑥) ∈ 𝐷𝑓

𝐷𝑔: 𝑥 ≠ −4 𝑔(𝑥) ≠ 5

𝑥−1

𝑥+4=

5

1

5𝑥 + 20 = 𝑥 − 1 4𝑥 = −21

𝑥 =−21

4

𝑥 = −5.25 𝑔(−5.25) = 5 𝑥 ≠ −5.25 𝐷𝑓𝑔 = (−∞, −5.25) ∪ (−5.25, −4) ∪ (−4, ∞)

45. 𝐷𝑓𝑔 = (−∞, −5.25) ∪ (−5.25, −4) ∪ (−4, ∞)

46. Let 𝑓(𝑥) =

𝑥+2

𝑥−5 and 𝑔(𝑥) =

𝑥−1

𝑥+4, find the

domain of (𝑔 ∘ 𝑓)(𝑥) 𝑥 ∈ 𝐷𝑓 and 𝑓(𝑥) ∈ 𝐷𝑔

𝑥 ≠ 5 𝑓(𝑥) ≠ −4

𝑥 + 2

𝑥 − 5=

−4

1

𝑥 + 2 = −4𝑥 + 20 5𝑥 = 18

𝑥 =18

5

𝑥 = 3.6 𝑓(3.6) = −4 𝑥 ≠ 3.6 𝐷𝑓∘𝑔 = (−∞, 3.6) ∪ (3.6,5) ∪ (5, ∞)

46. 𝐷𝑓∘𝑔 = (−∞, 3.6) ∪ (3.6,5) ∪ (5, ∞)

47. Let 𝑓(𝑥) = {𝑥2 + 1 𝑥 < 1

2𝑥 − 1 𝑥 ≥ 1

I. 𝑓(−2) = 5

II. 𝑓(1) = 1

III. 𝑓(4) = 7

48. Let 𝑓(𝑥) = 10 − 3𝑥

I. 𝑓(𝑥 + 2) = −3𝑥 + 4

II. 𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ= −3

48.

49. Let 𝑓(𝑥) = 𝑥2 − 11 and 𝑔(𝑥) = 5𝑥 + 4

I. (𝑓 ∘ 𝑔)(𝑥) = 25𝑥2 + 18𝑥 + 5 II. (𝑔 ∘ 𝑓)(𝑥) = 5𝑥2 − 51

49.

50.

50. ℎ(6) = 4 ℎ(5) = 6 ℎ(3) = 3 ℎ(5) = 5

51. Find the inverse of the relation {(−5, −1), (0, 4), (2, 6), (5, 9)} 51. {(−1, −5), (4, 0), (6, 2), (9, 5)}

52. Find the inverse of 𝑓(𝑥) =

1

𝑥−2

𝑦 =1

𝑥 − 2

𝑥 =1

𝑦 − 2

𝑥(𝑦 − 2) = 1

𝑦 − 2 =1

𝑥

𝑦 =1

𝑥+ 2

𝑓−1(𝑥) =1

𝑥+ 2

52. 𝑓−1(𝑥) =

1

𝑥+ 2

53.)

54.)

55.) It passes the vertical line test. It is a function It fails the horizontal line test. It is not a 1-1 function