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Math 121 Final Review

Example 1. Is the set of ordered pairs f = {(2, 1), (3, 1), (5, 1), (10, 1)} a function?

Solution

Yes each input goes to exactly one output.

Example 2. Determine if the graph below represents a function. No points will be given without anexplanation.

x

f(x)

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

Solution

No it does not pass the vertical line test

Example 3. What is the domain of f(x) =√x− 3. Write the answer in interval notation.

Solution

The domain is the set where the radicand is positive. x− 3 ≥ 0 ⇐⇒ x ≥ 3. This gives [3,∞)

Example 4. Assume that f(x) is a linear function and complete the table.

x -2 -1 0 1 2f(x) 9 4 -1 -6 -11

Solution

Since the function is linear is must have constant slope. m =−6− 4

1− (−1)=−10

2=−5

1=

∆y

∆x. So

a change of 1 unit for x produces a change of −5 units in y

Example 5. A linear function f(x) is given in the graph below. Find the slope and y intercept of f(x).

Intermediate Algebra Math 121 Intermediate Algebra


x

f(x)

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

∆x

∆y

Solution

The slope can be read off the graph m =∆y

∆x=−2

3. The y−intercept is can also be read off the

graph as (0, 1). This gives the equation of the line as y = mx + b =⇒ y = −2

3x + 1

Example 6. In your own words, explain what the symbol ∆t means.

Solution

The greek letter capitol delta means “The change in” so this is the change in t.

Example 7. True or False. Every constant function is a linear function.

Solution

True. Constant functions are linear with slope m = 0

Example 8. What is the slope of a vertical line?

Solution

Undefined

Example 9. Is the following function linear f(x) = 5− 3x, if so what are m and b?

Solution

The function is linear f(x) = mx + b. The value of m = −3 and b = 5

Example 10. True or False. Two lines are perpendicular if m1 ·m2 = −1



Solution

True

Example 11. Find the slope intercept form of the line that contains the points (−3,−7) and (0, 5)

Solution

m =5− (−7)

0− (−3)=

12

3= 4. The y−intercept is at (0, b) and so b = 5. This gives

y = 4x + 5

Example 12. A 4-GB iPod can hold about 1000 songs, whereas a 60-GB iPod can store 15,000 songs.

(a) Find a formula that calculates the number of songs, S, that can be stored on x gigabytes.

Solution

m =15000− 1000

60− 4= 250. This gives y = 250x + b. Using the first point gives

1000 = 250(4) + b ⇐⇒ b = 0. y = 250x

(b)How many songs can be stored on a 30-GB iPod?

Solution

y = 250(30) = 7500

Example 13. True or False. The perimeter formula for a rectangle is P = l · w

Solution

False. This is the area formula the perimeter is P = 2l + 2w

Example 14. True or False. 637% as a decimal is 6.37

Solution

True. To change a percent to a decimal move the decimal two places to the left.

Example 15. How much water should be added to 4 gallons of a 5% herbicide solution to dilute it toa 3.2% herbicide solution?

Solution



Percent herbicide Decimal Volume Mass5% solution .05 4 .05 · 4 = 0.20% (water) 0 x 0 · x = 0

3.2% 0.032 x + 4 0.032(x + 4)Equating the mass column gives

0.2 + 0 = 0.032(x + 4) ⇐⇒ 0.2 = 0.032x + .128 ⇐⇒ 0.72 = 0.032x ⇐⇒ x =0.72

0.032= 2.25

So we need to add 2.25 gallons of water.

Example 16. A student takes out two student loans, one at 6% annual interest and the other at 4%annual interest. The total amount of the two loans is $5000 and the total interest after 1 year is $244.Find the amount of each loan.

Solution

I = P r t0.06x = x 0.06 1

0.04(5000− x) = 5000− x 0.04 1244 = 5000 NA 1

Using the interest column gives

0.06x + 0.04(5000− x) = 244 ⇐⇒ 0.02x + 200 = 244 ⇐⇒ 0.02x = 44 =⇒ x =44

0.02= 2200

So there was $2200 in the 6% account and $2800 in the 4% account

Example 17. Two cars are traveling in opposite lanes on a freeway. Two and a half hours after theymeet they are 355 miles apart. If one car is traveling 6 miles per hours faster than the other car, findthe speed of each car.

Solution

d = r tCar 1 2.5x = x 2.5Car 2 2.5(x + 6) = x + 3 2.5Total 355 = NA 2.5

Using the distance column gives2.5x + 2.5(x + 6) = 355 ⇐⇒ 5x + 15 = 355 ⇐⇒ 5x = 340 =⇒ x = 68

One car was traveling at 68mph and the other car at 74mph

Example 18. Is x = 6 a solution to the inequality 2x− 5 ≤ 0

Solution

2(6)− 5 = 12− 5 = 7 � 0 No it is not a solution.

Example 19. Is x = 1 a solution to x < −2 or x > 2



Solution

No it does not check in either inequality.

Example 20. Solve the inequality1

3x ≥ 6

Solution

3 · 1

3x ≥ 3 · 6 ⇐⇒ x ≥ 18

Example 21. Solve the inequality 5− 2x ≤ 17

Solution

5− 2x ≤ 17 ⇐⇒ −2x ≤ 12 ⇐⇒ x ≥ −6

Example 22. Express in interval notation. x > 5

Solution

(5,∞)

Example 23. Write the compound inequality using interval notation. −2 ≤ x ≤ 3

Solution

[-2,3]

Example 24. Write the inequality in interval notation. x < −2 or x ≥ 5

Solution

(−∞,−2) ∪ [5,∞)

Example 25. Write the compound inequality as an absolute value inequality. y < −3 or y > 3

Solution

|y| > 3

Example 26. x = −3 is a solution to |x| = 3



Solution

True. | − 3| = 3

Example 27. Solve the inequality |x| > 5

Solution

x < −5 or x > 5

Example 28. How many times does the graph of y = |x + 2| intersect the graph of y = 1

Solution

Making a table and plotting points gives

x

f(x)

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−6

−5

−4

−3

−2

−1

1

2

3

4

5

6

x −6 −4 −2 0 2 4f(x) 4 2 0 2 4 6

The graph intersects twice.

Example 29. Solve |4x− 7| > 5

Solution

This is an “or” type compound inequality. This gives:4x− 7 < −5 or 2x− 7 > 5

4x < 2 4x > 12

x <1

2x > 3

Example 30. Solve the inequality |5− 3x| − 1 > 11



Solution

|5− 3x| − 1 > 11 ⇐⇒ |5− 3x| > 12 Case II: 5− 3x < −12 or 5− 3x > 12

Solving the first gives 5− 3x < −12 ⇐⇒ −3x < −17 ⇐⇒ x >17

3

Solving the second gives 5− 3x > 12 ⇐⇒ −3x > 7 ⇐⇒ x < −7

3

Example 31. For the equations 2x − y = 6 and 2x − y = 12 Classify as Inconsistent, ConsistentDependent, or Consistent Independent.

Solution

Inconsistent

Example 32. Determine if (−2, 7) is a solution to the linear system 5x − 6y = −52 − 3x − 4y =−34

Solution

Eqn1: 5(−2)− 6(7) = −10− 42 = −52 Eqn2: −3(−2)− 4(7) = 6− 28 = −22 6= −34So the point is not a solution.

Example 33. Solve the linear system graphically 2x− y = 4 3x + y = −9

x

f(x)

−6 −4 −2 2 4 6

−6

−4

−2

2

4

6

Solution

The point of intersection is (−1,−6)

Example 34. True or False. A system of equations can have exactly two solutions.

Solution



False.

Example 35. True or False. A system of parallel lines has exactly one solution.

Solution

False. The system has no solutions.

Example 36. True or False. If elimination gives 4 = 0. The solution to the linear system is (4, 0)

Solution

False the system has no solutions.

Example 37. Solve the linear system by substitution. 2x− 3y = −28 y = 3x

Solution

2x− 3(3x) = −28 ⇐⇒ −7x = −28 ⇐⇒ x = 4y = 3(4) = 12 =⇒ (4, 12)

Example 38. Solve the linear system by elimination 3x− 7y = 5 3x + 7y = 7

Solution

Multiplying the 2nd equation by −1 and adding gives3x + 7y = 5−3x− 7y = −7

0 = −2This equation is false. So the system has no solution.

Example 39. Solve the linear system 3x− 5y = 6 − 6x + 10y = −12

Solution

Multiplying the 1st equation by 2 and adding gives6x− 10y = 12−6x + 10y = −12

0 = 0This equation is True. So the system has an infinite number solutions. {(x, y)|3x− 5y = 6}

Example 40. A boat travels downstream 150 miles in 5 hours. The return trip takes 7.5 hours. Findthe speed of the boat without a current and the speed of the current.

Solution



Let B be the speed of the boat and C be the current. Using the model d = rt

d r tUpstream 150 B − C 7.5

Downstream 150 B + C 5

This gives the system of equations

{150 = (B − C)(7.5)

150 = (B + C)(5)

Dividing equation 1 by 5 and equation 2 by 7.5 will give the variable C equal but oppositecoefficients.{20 = B − C

30 = B + C

Adding the equations gives

50 = 2B =⇒ B =50

2= 25 =⇒ C = 5

The speed of the boat is 25 miles per hour and the speed of the current is 5 miles per hour.

Example 41. A chemist needs 30 liters of a 4% acid solution. She only has an 2% and 70% solution.How many liters of each should she mix to obtain the 4% solution? Use a system of equations tosolve.

Solution

Let x be the number of liters of the 2% solution and y be the number of liters of the 70%solution. The equation for the volume of the acid it x + y = 30. The equation for the part of the

solution that is acid is .02x + .70y = .04(30) Multiplying this equation by 100 to clear thedecimals gives the system,{

x + y = 30

2x + 70y = 120

Multiplying the first equation by −2 and adding it to the second equation gives−2x + −2y = −60

2x + 70y = 12068y = 60

This gives that y =60

68=

15

17. Substituting this into the first equation gives

15

17+ y = 30 ⇐⇒ y = 30− 15

17=

30 · 17− 15

17=

495

17

The chemist needs495

17liters of 2% acid and

15

17liters of 70% acid.

Example 42. Does the opposite of x2 + 1 equal −x2 + 1

Solution

No, it is −(x2 + 1) = −x2 − 1



Example 43. Could the graph of a polynomial function be a line?

Solution

Yes a degree one polynomial is linear

Example 44. If f(x) = 2x2 − 4x, then f(3)

Solution

f(3) = 2(3)2 − 4(3) = 18− 12 = 6

Example 45. Is3

x− 2a monomial?

Solution

No it has a variable with a positive exponent in the denominator.

Example 46. Write a monomial that represents the described quantity. The area of a square with sidelength equal to 2x

Solution

2x

2x The area of the rectangle is (2x)(2x) = 4x2

Example 47. The number of members in a marching band that has x rows with y people in each row.

Solution

x · y

Example 48. Identify the degree and leading coefficient. 3x6

Solution

coefficient= 3 degree= 6

Example 49. Identify the degree and leading coefficient. x3y4

Solution

coefficient= 1 degree= 3 + 4 = 7



Example 50. Identify the degree and leading coefficient. 5− x + 3x2 − 2

5x3

Solution

coefficient= −2

5degree= 3

Example 51. Identify the degree and leading coefficient. x2 − 2x + 1

Solution

coefficient= 1 degree= 2

Example 52. Simplify 5x + 4y

Solution

Can’t be simplified

Example 53. Simplify (3x3 + 2x2 − 4) + (−3x2 + x + 4)

Solution

= 3x3 − x2 + x

Example 54. Simplify (2x2 + 5x− 3)− (−2x2 + 5x− 3)

Solution

= 4x2

Use f(x) = 5x− 2 and g(x) = x2 − 4 to answer the following questions.

Example 55. (f + g)(x)

Solution

(5x− 2) + (x2 − 4) = x2 + 5x− 6

Example 56. (f + g)(2)

Solution

22 + 5(2)− 6 = 8

Example 57. (f − g)(x)

Solution

(5x− 2)− (x2 − 4) = −x2 + 5x + 2



Example 58. (f − g)(−2)

Solution

−(−2)2 + 5(−2) + 2 = −12

Example 59. Does (x + 1)2 = x2 + 1? Explain.

Solution

No (x + 1)(x + 1) = x2 + 2x + 1

Example 60. Does x3(x2 − x4) = x6 − x12

Solution

No x3+2 − x3+4 = x5 − x7

Example 61. Multiply (2x) · (4x3)

Solution

8x4

Example 62. Multiply −3x(4− 2x)

Solution

6x2 − 12x

Example 63. Multiply −(9− x2)

Solution

x2 − 9

Example 64. Multiply (x + 3)(x− 2)

Solution

x2 − 2x + 3x− 6 = x2 + x− 6

Example 65. Multiply (x− 2)(x2 + 2x + 4)

Solution

x3 + 2x2 + 4x− 2x2 − 4x− 8 = x3 − 8

Example 66. Multiply (6x− 5)2



Solution

(6x− 5)(6x− 5) = 36x2 − 30x− 30x + 25 = 36x2 − 60x + 25

Use f(x) = x2 and g(x) = x2 − 9 to answer the following questions.

Example 67. (fg)(x)

Solution

x2(x2 − 9) = x4 − 9x2

Example 68. (fg)(1)

Solution

14 − 9(1)2 = 1− 9 = −8

Example 69. (fg)(2)

Solution

24 − 9(2)2 = 16− 36 = −20

Example 70. (fg)(3)

Solution

34 − 9(3)2 = 81− 81 = 0

Example 71. If x(x− 3) = 0, what can be said about either x or x− 3?

Solution

By the zero product rule each factor must equal zero.x = 0 or x− 3 = 0

Example 72. 2x is the GCF of 4x3 − 12x2

Solution

No it is 4x2

Example 73. If a · b = 2, then either a = 2 or b = 2

Solution

False. There is not a 2 product principle. There is only a zero product principle.

Example 74. Factor 32− 16x



Solution

= −16(x− 2)

Example 75. Factor 3r3 − 24r6

Solution

= −3r3(8r3 − 1) The 2nd factor is the difference of cubes see 5.5

Example 76. Factor 17x4 − 2x2 − x

Solution

= x(17x3 − 2x− 1)

Example 77. Solve (x− 1)(x + 2) = 0

Solution

By the zero product rulex− 1 = 0 =⇒ x = 1 or x + 2 = 0 =⇒ x = −2

Example 78. Solve 3x(x− 7) = 0

Solution

By the zero product rule3x = 0 =⇒ x = 0 or x− 7 = 0 =⇒ x = 7

Example 79. Solve (x + 1)(x− 2)(x + 3) = 0

Solution

By the zero product rule(x + 1) = 0 =⇒ x = −1 or x− 2 = 0 =⇒ x = 2 or x + 3 = 0 =⇒ x = −3

Example 80. Solve x2 − 2x = 0

Solution

By factoring and the zero product rulex(x− 2) = 0 =⇒ x = 0 or x− 2 = 0 =⇒ x = 2

Example 81. Solve 5x2 − x = 0

Solution

By factoring and the zero product rule

=⇒ x(5x− 1) = 0 =⇒ x = 0 or 5x− 1 = 0 =⇒ x =1

5



Example 82. Solve 2x(x + 2) + 3(x + 2) = 0

Solution

By factoring and the zero product rule

(2x + 3)(x + 2) = 0 =⇒ 2x + 3 = 0 =⇒ x = −3

2or x + 2 = 0 =⇒ x = −2

Example 83. The height h in feet reached by a golf ball after t seconds is given by h(t) = −16t2 + 96t.After how many seconds does the golf ball strike the ground?

Solution

−16t2 + 96t = 0 ⇐⇒ −16t(t− 6) = 0 This gives the two solutions t = 0 or t = 6. The ball hitsthe ground after 6 seconds.

Example 84. Factor x2 + 8x + 12

Solution

= (x + 2)(x + 6)

Example 85. Factor x2 − 8x + 12

Solution

= (x− 2)(x− 6)

Example 86. Factor x2 − 4x− 12

Solution

= (x− 6)(x + 2)

Example 87. Factor x2 − 18x + 72

Solution

= (x− 6)(x− 12)

Example 88. Factor 2x2 + 7x + 3

Solution

= (2x + 1)(x + 3)

Example 89. Factor 10x2 + 3x + 1



Solution

Does not factor

Example 90. Factor 6x2 − 13x + 6

Solution

= (2x− 3)(3x− 2)

Example 91. Factor 1 + x− 2x2

Solution

= −[2x2 − x− 1] = −(2x + 1)(x− 1)

Example 92. Factor 5x3 + x2 − 6x

Solution

=x[5x2 + x− 6] = x(5x + 6)(x− 1)

Example 93. Factor x4 − x3 − 2x2

Solution

= x2[x2 − x− 2] = x2(x− 2)(x + 1)

Example 94. Factor 60x3 + 230x2 − 40x

Solution

= 10x[6x2 + 23x− 4] = 10x(6x− 1)(x + 4)

Example 95. Factor 30x4 + 3x3 − 9x2

Solution

= 3x2[10x2 + x− 3] = 3x2(5x + 3)(2x− 1)

Example 96. Factor x2 − 144

Solution

= (x− 12)(x + 12)

Example 97. Factor 36x2 − 100



Solution

= 4[9x2 − 25] = 4(3x− 5)(3x + 5)

Example 98. Factor x2 + 9

Solution

Does not factor

Example 99. Factor x3 − 125

Solution

= (x− 5)(x2 + 5x + 25)

Example 100. Factor 125x6 + 1

Solution

= (5x2 + 1)(25x4 − 5x + 1)

Example 101. Factor a8 − b8

Solution

= (a4 − b4)(a4 + b4) = (a2 − b2)(a2 + b2)(a4 + b4) = (a− b)(a + b)(a2 + b2)(a4 + b4)

Example 102. Solve x2 − 2x− 15 = 0

Solution

Since a = 1 the quadratic can be factored using the pq method. Factors of c that add up to b areneeded.

We see that (−5) + (3) = −2 and (−5)(3) = −15. This gives

(x− 5)(x + 3)

Isolating x in each factor gives the solutions x = 5 and x = −3

Example 103. Solve x2 − 4x− 32 = 0

Solution

Since a = 1 the quadratic can be factored using the pq method. Factors of c that add up to b areneeded.

We see that (−8) + (4) = −4 and (−8)(4) = −32. This gives

(x− 8)(x + 4)



Isolating x in each factor gives the solutions x = 8 and x = −4

Example 104. Solve 8x2 − 30x + 25 = 0

Solution

Since a 6= 1 the quadratic can be factored using the ac method. Multiplying a and c togethergives 200

Factors of 200 that add up to −30 are needed.We see that (−20) + (−10) = −30 and (−20)(−10) = 200

Splitting the b term up gives

8x2 − 20x− 10x + 25

Factoring out the common term from each group yields

4x(2x− 5)− 5(2x− 5)

Factoring out the new common factor (2x− 5) from each term gives

(2x− 5)(4x− 5)

Isolating x in each factor gives the solutions x =5

2and x =

5

4

Example 105. Solve 12x2 + 32x + 21 = 0

Solution

Since a 6= 1 the quadratic can be factored using the ac method. Multiplying a and c togethergives 252

Factors of 252 that add up to 32 are needed.We see that (14) + (18) = 32 and (14)(18) = 252

Splitting the b term up gives

12x2 + 14x + 18x + 21

Factoring out the common term from each group yields

2x(6x + 7) + 3(6x + 7)

Factoring out the new common factor (6x + 7) from each term gives

(6x + 7)(2x + 3)

Isolating x in each factor gives the solutions x = −7

6and x = −3

2

Example 106. Identify p(x), q(x), find the domain and vertical asymptote(s) of f(x) =x

x− 2



Solution

A rational function has the form f(x) =p(x)

q(x). This gives p(x) = x and q(x) = x− 2. To find the

domain the zero’s of q(x) must be excluded. x− 2 = 0 ⇐⇒ x = 2. The domain is {x|, x 6= 2}.The vertical asymptotes are the zeros of q(x). This gives x = 2.

Example 107. Find the domain of f(x) =x− 3

x2 − 2x− 3.

Solution

The zeros of the denominator (q(x)) must be excluded from the domain.q(x) = x2 − 2x− 3 ⇐⇒ x2 − 2x− 3 = 0 ⇐⇒ (x− 3)(x + 1) = 0. So the values x = 3 and

x = −1 must be excluded. This gives {x|x 6= 3 and x 6= −1}

Example 108. Graph the equation f(x) =x− 2

x + 2Please label the vertical asymptote.

x

y

−10 −8 −6 −4 −2 2 4 6 8 10

−10

−8

−6

−4

−2

2

4

6

8

10←− Vertical Asymptote x = −2

f(−3) =−3− 2

−3 + 2= 5 ⇐⇒ (−3, 5)

f(−5) =−5− 2

−5 + 2=

7

3⇐⇒ (−5,

7

3)

f(−7) =−7− 2

−7 + 2=

9

5⇐⇒ (−7,

9

5)

f(−9) =−9− 2

−9 + 2=

11

7⇐⇒ (−9,

11

7)

f(−1) =−1− 2

−1 + 2= −3 ⇐⇒ (−1,−3)

f(1) =1− 2

1 + 2= −1

3⇐⇒ (1,−1

3)

f(3) =3− 2

3 + 2=

1

5⇐⇒ (3,

1

5)

f(5) =5− 2

5 + 2=

3

7⇐⇒ (5,

3

7)

f(7) =7− 2

7 + 2=

5

9⇐⇒ (7,

5

9)

f(9) =9− 2

9 + 2=

7

11⇐⇒ (9,

7

11)

Example 109. If f(x) = x + 2 and g(x) = x2 + 1, find

(f

g

)(−2) and

(f

g

)(2)

Solution

(f

g

)(−2) =

f(−2)

g(−2)=−2 + 2

(−2)2 + 1=

0

5= 0

(f

g

)(2) =

f(2)

g(2)=

2 + 2

(2)2 + 1=

4

5

Example 110. Simplify: a)x2 + 1

x2 − 1· x− 1

x + 1b)

x2 − 4

x2 + x− 2÷ x− 2

x− 1



Solution

Factoring gives

a)x2 + 1

(x− 1)(x + 1)· x− 1

x + 1=

x2 + 1

(x + 1)2

b)(x− 2)(x + 2)

(x + 2)(x− 1)÷ x− 2

x− 1=

(x− 2)(x + 2)

(x + 2)(x− 1)· x− 1

x− 2= 1

Example 111. Simplify18t3

6tand

a2 − b2

a− b

Solution

18t3

6t=

6t

6t· 3t2

1= 3t2

a2 − b2

a− b=

(a− b)(a + b)

a− b= a + b

Example 112. Multiply5a

4· 12

5aand

2x + 4

x + 1· x

2 + 3x + 2

4x + 2

Solution

5a

4· 12

5a=

60a

20a= 3

2x + 4

x + 1· x

2 + 3x + 2

4x + 2=

2(x + 2)

x + 1· (x + 2)(x + 1)

2(2x + 1)=

(x + 2)2

(2x + 1)

Example 113. Divide3x

2÷ 2x

5and

x2 − 4

x2 + x− 2÷ x− 2

x− 1

Solution

3x

2÷ 2x

5=

3x

2· 5

2x=

15x

4x=

15

4

x2 − 4

x2 + x− 2÷ x− 2

x− 1=

x2 − 4

x2 + x− 2· x− 1

x− 2=

(x− 2)(x + 2)

(x + 2)(x− 1)· x− 1

x− 2= 1

Example 114. Find the LCM of 2x2 − 2x and 8x2.

Solution

Factor each expression to get 2x(x− 1) and 23 · x2. Taking the factor with the largest exponentgives the LCD 23x(x− 1) = 8x2(x− 1)

Example 115. Simplify3x2

x2 − 4− 2x2 + 4

x2 − 4

Solution



3x2

x2 − 4− x2 + 4

x2 − 4=

3x2 − (2x2 + 4)

x2 − 4=

x2 − 4

x2 − 4= 1

Example 116. Simplify4x

x2 − 9+

8

x− 3

Solution

4x

x2 − 9+

8

x− 3=

4x

(x− 3)(x + 3)+

8(x + 3)

(x− 3)(x + 3)=

12x + 24

(x + 3)(x− 3)

Example 117. Simplify: a)3

x2 − 1− 1

x + 1b)

1

2(x− 3)+

1

2(x− 3)

Solution

a)3

(x− 1)(x + 1)− 1

x + 1·(x− 1

x− 1

)=

3

(x + 1)(x− 1)− x− 1

(x + 1)(x− 1)=

−x + 4

(x− 1)(x + 1)

b)1

2(x− 3)+

1

2(x− 3)=

2

2(x− 3)=

1

(x− 3)

Example 118. Solvex2 − 4x− 5

x− 1+

8

x− 1= 0

Solution

Multiply by the LCD to clear the fractions, this gives

x2 − 4x− 5

x− 1· (x− 1) +

8

x− 1· (x− 1) = 0 · (x− 1)

x2 − 4x− 5 + 8 = 0 ⇐⇒ x2 − 4x + 3 = 0 =⇒ (x− 3)(x− 1) = 0

So the two possible solutions are x = 3 and x = 1. The solution x = 1 must be rejected, becauseit leads to division by 0.

Example 119. Solve the equation3

x− 2= x

Solution

3

x− 2· (x− 2) = x(x− 2) ⇐⇒ 3 = x2 − 2x ⇐⇒ x2 − 2x− 3 = 0 ⇐⇒ (x− 3)(x + 1) = 0 So

the solutions are x = 3 and x = −1

Example 120. Solve4

5x= − 1

x + 1

Solution



Multiply by the LCD 5x(x + 1) to clear the fractions, this gives

4

5x= − 1

x + 1⇐⇒ 4

5x· 5x(x + 1) = − 1

x + 1· 5x(x + 1)

4(x + 1) = −5x ⇐⇒ 9x = −4 =⇒ x = −4

9

Example 121. Solve the rational equation3

x− 1=

6

x + 4.

Solution

The possible solution x = 1 and x = −4 must be excluded, it is not in the domain of theequation. The fractions can be cleared by multiply by the LCD (x− 1)(x + 4)

3

x− 1· (x + 4)(x− 1) =

6

x + 4· (x + 4)(x− 1) ⇐⇒ 3(x + 4) = 6(x− 1)

3x + 12 = 6x− 6 ⇐⇒ 18 = 3x ⇐⇒ x = 6 Since x = 6 is not an excluded solution, it willsatisfy the equation.

Example 122. At a raffle each person is given 1 ticket and one ticket is drawn to determine the winner.

The probability P (x) of winning is given by P (x) = 1 − x− 1

x. If your probability of winning is 1%,

how many tickets are in the raffle?

Solution

Setting P (x) = 0.01 =1

100gives

1

100= 1− x− 1

xMultiplying by the LCD and simplifying gives

x = 100x− 100(x− 1) ⇐⇒ x = 100 So there are 100 tickets in the raffle.

Example 123. A push lawn mower can mow a lawn in 5 hours. A riding lawn mower can mow a lawnin 3 hours. How long will it take if both mowers are used? Please round the answer to the nearesthundredth.

Solution

The portion of the job the push mower does in one hour is1

5and the riding mower’s portion in

one hour is1

3. Multiplying each rate by the number of hours,t, worked gives the amount of work

done in t hours..

t

5+

t

3= 1

Clearing the fractions and collecting like terms gives

3t + 5t = 15 ⇐⇒ 8t = 15 ⇐⇒ t ≈ 1.88

If both mowers are used it will take about 1.88 hours



Example 124. Simplify

x2y2x3y

Solution

x2y2x3y

=x

2y· 3y

2x=

3

4

Example 125. Simplify the equation R =1

1R1

+ 1R2

Solution

Multiply the numerator and denominator by R1R2 this gives R =R1R2

R1R2

R1+ R1R2

R2

=R1R2

R2 + R1

Example 126. In an electric circuit, when resistors (R1 and R2) are wired in parallel, the combine

resistance R is given by the formula1

R=

1

R1+

1

R2. Find the combined resistance if R1 = 5 ohms and

R2 = 3 ohms.

Solution

Plugging in gives1

R=

1

5+

1

3=

3

15+

5

15=

8

15⇐⇒ R =

15

8


1p−11

p−1 + 2

Solution

1p−11

p−1 + 2=

1p−1

1p−1 + 2(p−1)

p−1

=

1p−12p−1p−1

=1

p− 1· p− 1

2p− 1=

1

2p− 1

Example 128. y varies directly as x. If y = 6 when x = 3, find y when x = 7

Solution

y = kx ⇐⇒ 6 = 3k ⇐⇒ k = 2 =⇒ y = 2x =⇒ y = 2(7) = 14

Example 129. Three hundred large-mouth bass are tagged and released into a lake. Later, a sampleof 112 large-mouth bass contains 17 tagged bass. Estimate the large-mouth bass population to thenearest hundred.



Solution

x

300=

112

17=⇒ x =

112 · 300

17≈ 1976.47

Rounding to the nearest hundred gives 2000 large-mouth bass.

Example 130. The distance d that it takes a car to stop is directly proportional to the square of itsspeed v. If a car moving 30 miles per hour takes 145 feet to stop, how many feet will a car traveling 45miles per hour take to stop? Round the final answer to the nearest foot.

Solution

The equation of variation is d = kv2. Solving for the proportionality constant gives

145 = k(30)2 ⇐⇒ k =145

302≈ 0.1611

Using the unrounded value of k from the calculator gives d = (0.1611)(45)2 ≈ 326 The car willtake about 326 feet to stop.

Example 131. Hooke’s Law states that the restoring force of a spring is directly proportional to theamount the spring is stretched from its natural length. If a spring is stretched 2 meters, the restoringforce is 10 Newtons. How much would a force of 30 Newtons stretch this spring?

Solution

F = kx where F is the force and x is the stretch. 10 = k(2) ⇐⇒ k = 5. Plugging k back intothe equation gives. F = 5x If the force is 30N then 30 = 5x ⇐⇒ x = 6.

Example 132. Divide50x4 + 15x2 + 100x

25x

Solution

50x4 + 15x2 + 100x

25x=

50x4

25x+

15x2

25x+

100x

2x3= x2 +

3

5x + 4

Example 133. Divide (Use long division)4x2 − x− 18

x + 2

Solution

4x− 9

Example 134. Simplify√x2

Solution



√x2 = |x|

Example 135. Simplify√−36

Solution

Not a real number.

Example 136. Simplify 3√−64

Solution

Since (−4)3 = −64 3√−64 = −4

Example 137. True or False7√

7 = 1

Solution

False.

Example 138. Write as a rational exponent n√xm

Solution

xmn

Example 139. Simplify 9−12

Solution

9−12 =

1

912

=1√9

=1

3

Example 140. Use positive rational exponents to simplify the expression. Assume all variables are

positive.√y3 · 3

√y2

Solution

y32 · y 2

3 = y32+

23 Adding the exponents gives

2

3+

3

2=

4 + 9

6=

13

6This gives

y136 = 6

√y13 = y2

√y

Example 141. Simplify3√

4 ·√

16



Solution

43√

4


√x

2·√

x

8

Solution √x

2·√

x

8=

√x2

16=

x

4


√4xy2√x

Solution √4xy2√x

=

√4xy2

x=√

4y2 = 2y


√x2 − 4x + 4√

x− 2

Solution√x2 − 4x + 4√

x− 2=

√x2 − 4x + 4

x− 2=

√(x− 2)2

x− 2=√x− 2

Example 145. A rectangle has a side length of√

32 units and a height of√

8 units. Find the area andperimeter of the rectangle.

Solution

The perimeter of a rectangle is P = 2l + 2w and the area is A = lw. Evaluating each formulagives:

P = 2(√

32) + 2(√

8) = 8√

2 + 4√

2 = 12√

2

A = (√

32)(√

8) = (4√

2)(2√

2) = 16

Example 146. Simplify the expression. 2√

3 + 2√

12 + 3√

27

Solution

Simplifying each radical gives 2√

3 + 2√

4 · 3 + 3√

9 · 3 = 2√

3 + 4√

3 + 9√

3 = 15√

3

Example 147. Simplify the expression. (8 +√x)(2−

√x)



Solution

Using the distributive law gives 16− 8√x + 2

√x− x = −x− 6

√x + 16

Example 148. Rationalize the denominator.

√2

2−√

5

Solution

Multiplying the numerator and denominator by the conjugate of the denominator gives

√2

2−√

5· 2 +

√5

2 +√

5=

2√

2−√

10

4− 5=

2√

2−√

10

−1=√

10− 2√

2

Example 149. Evaluate the function f(x) =√x− 5 at the given values x = −2, x = 6, and x = 9.

Solution

f(−2) =√−2− 5 =

√7i. So it is not a real number. f(6) =

√6− 5 =

√1 = 1.

f(9) =√

9− 5 =√

4 = 2.

Example 150. Find the domain of the function g(x) =√

2x− 9. Write the answer in interval notation.

Solution

The domain of g(x) is the set of values such that the radicand is non-negative.

2x− 9 ≥ 0 ⇐⇒ 2x ≥ 9 ⇐⇒ x ≥ 9

2. The domain is the set

[9

2,∞)

Use the rectangle below to answer the following questions.√

18x cm

√8x cm

Example 151. Find the exact perimeter of the rectangle. Hint: P = 2l + 2w

Solution

P = 2√

18x + 2√

8x = 2√

9 · 2x + 2√

4 · 2xP = 6

√2x + 4

√2x = 10

√2x cm

Example 152. Find the exact area of the rectangle. Hint: A = lw



Solution

A =√

18x ·√

8x =√

144x2 = 12x cm2

Example 153. The time for an object to fall x feet is given by t =

√x

4. If a ball is dropped from a

building with a height of 256 feet, how long will it take for the ball to hit the ground?

Solution

t =

√256

4=

16

4= 4. The object will take 4 seconds to hit the ground.

Example 154. The time for an object to fall x feet is given by t =

√x

4. If a ball is dropped from a

building and falls for 7 seconds, how tall is the building?

Solution

7 =

√x

4⇐⇒ 28 =

√x ⇐⇒ x = 282 = 784. The building is 784 feet tall.

Example 155. Find the length of the missing side of the right triangle.

10 units

8 units

Solution

Using the Pythagorean Theorem with a = 8, b =? and c = 10 gives82 + x2 = 102 ⇐⇒ x2 = 36 =⇒ x = 6

Use the following to answer Exercise 156 and 157.Johannes Kepler discovered a relationship between a planet’s distance D from the sun and the time Tit takes to orbit the sun. This formula is T =

√D3 where T is in Earth years and D = 1 corresponds

to the distance between the Earth and the sun.

Example 156. Write the formula with rational exponents.

Solution

T = D32

Example 157. The planet Neptune is 30 times from the sun than Earth (D = 30). Use a calculator toestimate the number of years required for Neptune to orbit the sun. Round your solution to the nearest



whole number.

Solution

T = 3032 ≈ 164 rounded to the nearest year.

Example 158. Solve√

3x + 4 = 8

Solution

Squaring both sides gives 3x + 4 = 64 =⇒ 3x = 60 =⇒ x = 20Checking the solution in the equation gives

√3(20) + 4 =

√60 + 4 =

√64 = 8

The solution to the equation is x = 20

Example 159. Simplify√−36

Solution

The solution is an imaginary number.√−36 =

√36i = 6i

Example 160. Identify the real and imaginary part of 3− 2i

Solution

The real part is a = 3 and the imaginary part is b = −2

Example 161. Simplify√−2√−18

Solution√

2i ·√

18i = i2√

36 = −1 · 6 = −6

Example 162. Multiply (3− 4i)(6 + 11i)

Solution

18 + 33i− 24i− 44i2 = 18 + 9i + 44 = 62 + 9i

Example 163. Simplify i785

Solution

4 divides into 785 196 with a remainder of 1. This gives i785 = i1 = i

Example 164. Find the conjugate of 4− 5i



Solution

4 + 5i

Example 165. Divide3i

4− 5i

Solution

3i

4− 5i· 4 + 5i

4 + 5i=

12i + 15i2

16 + 25=−15 + 12i

41= −15

41+

12

41i

8 Chapter 8

Example 166. Graph the equation f(x) = x2 + 2x− 8. Identify and label the vertex.

x

f(x)

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

Vertex

f(−5) = (−5)2 + 2(−5)− 8 = 7 ⇐⇒ (−5, 7)

f(−4) = (−4)2 + 2(−4)− 8 = 0 ⇐⇒ (−4, 0)

f(−3) = (−3)2 + 2(−3)− 8 = −5 ⇐⇒ (−3,−5)

f(−2) = (−2)2 + 2(−2)− 8 = −8 ⇐⇒ (−2,−8)

f(−1) = (−1)2 + 2(−1)− 8 = −9 ⇐⇒ (−1,−9)

f(0) = (0)2 + 2(0)− 8 = −8 ⇐⇒ (0,−8)

f(1) = (1)2 + 2(1)− 8 = −5 ⇐⇒ (1,−5)

f(2) = (2)2 + 2(2)− 8 = 0 ⇐⇒ (2, 0)

f(3) = (3)2 + 2(3)− 8 = 7 ⇐⇒ (3, 7)

Example 167. Find the vertex of f(x) = −16x2 + 4x + 2 and determine if the vertex is a maximumor a minimum value.

Solution

Since a < 0 the quadratic opens down and has a maximum value. The maximum value is located

at the vertex (h, k) and can be found using the vertex formula. h = − b

2aand k = f(h). Using

standard form f(x) = ax2 + bx + c =⇒ a = −16 and b = 4, this gives h = − 4

2(−16)=

1

8.

k = f

(1

8

)= −16

(1

8

)2

+ 4

(1

8

)+ 2 = −1

4+

1

2+ 2 =

9

4

Example 168. The fish population in a lake is given by F (t) = −1000t2 + 13000t+ 30000, where F isthe number of fish at time t, and t is measured in years since January 1, 2002, when the fish popula-tion was first estimated. Based on this model, on what day will the lake have the largest fish population?



Solution

Since the function is a quadratic, it will have its maximum at its vertex. Using the vertex

formula gives h = − 13000

2(−1000)=

13

2=6.5. The maximum population will occur six and one half

years from January 1, 2002. That will be July 1, 2008.

Example 169. A farmer, fencing a rectangular area for cattle, uses a barn as one side of the rect-angle. If the farmer has 1200 feet of fence, find the dimensions for the rectangle that maximizes the area.

Solution

The farmer only needs to put fence on three (perimeter) sides this gives P = 2x + y and the areais A = xy. The amount of fence is the perimeter of the 3 sided rectangular area. This gives

1200 = 2x + y ⇐⇒ y = 1200− 2x. Substituting this into the equation for area givesA = x(1200− 2x) = 2x(600− x). The maximum can be found by using the vertex formula asabove or by using symmetry. By symmetry the vertex must be at the midpoint of its zeros.

Since the zeros are x = 0 and x = 600 this gives. h =0 + 600

2= 300. The side perpendicular to

the river should be 300 feet and the side parallel to the river should be 600ft.

Example 170. Determine the vertex of f(x) = −3(x+ 4)2 − 3 and decide if the parabola opens up ordown.

Solution

Since the quadratic is in vertex form f(x) = a(x− h)2 + k =⇒ a = −3, h = −4 and k = −3.Since a < 0 the parabola opens down and the vertex is located at (h, k) ⇐⇒ (−4,−3)

Example 171. Write the equation of the parabola with a = −3 and vertex (−6, 3). Write your solutionin the form y = ax2 + bx + c

Solution

Using vertex form gives f(x) = −3(x + 6)2 + 3. Expanding to standard form givesf(x) = −3(x2 + 6x + 6x + 36) + 3 =⇒ f(x) = −3x2 − 36x− 105

Example 172. Write the equation of the quadratic in vertex form. Assume a = ±1



x

y

−2 0 2

−4

−2

2

4

(−1,−4)

(−3, 0) (1, 0)

Solution

Identifying (h, k) from the graph gives h = −1 and k = −4. Using the vertex from equation gives

f(x) = a(x− h)2 + k =⇒ f(x) = 1(x + 1)2 − 4 ⇐⇒ f(x) = (x + 1)2 − 4

Example 173. Solve (x− 5)2 = 11

Solution

Using the Square root principle gives x− 5 = ±√

11 =⇒ x = 5±√

11

Example 174. Solve by completing the square. x2 − 3x = 5

Solution

Taking half of the b term and squaring it gives

(−3

2

)2

=9

4and adding this to both sides gives

x2 − 3x +9

4= 5 +

9

4⇐⇒

(x− 3

2

)2

=29

4Using the square root property gives

x− 3

2=±√

29

2=⇒ x =

3±√

29

2

Example 175. Solve the quadratic 3x2 + 12x = 36

Solution

3x2 + 12x− 36 = 0 ⇐⇒ 3(x2 + 4x− 12) = 0 ⇐⇒ 3(x + 6)(x− 2) = 0 Using the zero productprinciple gives x + 6 = 0 or x− 2 = 0, so x = −6 or x = 2

Example 176. Solve 3x2 + 2x− 10 = 0

Solution

Identifying a,b, and c and using the quadratic formula gives



x =−b±

√b2 − 4ac

2a=−2±

√(2)2 − 4(3)(−10)

2(3)=−2±

√124

6

−2± 2√

31

6

Using Monomial division gives

−1

3± 1

3

√31

Example 177. Solve the equation t(t− 4) = −8

Solution

t(t− 4) = −8 ⇐⇒ t2 − 4t = −8 ⇐⇒ t2 − 4t + 8 = 0 The factors of eight are 8 = 1 · 8 = 2 · 4.Since neither pair sums to 4 the equation cannot be factored. It can be solved by using the

quadratic formula. Identifying a = 1, b = −4, and c = 8 gives

t =−(−4)±

√(−4)2 − 4(1)(8)

2(1)=

4±√−16

2= 2± 2i

Example 178. Use the discriminant to determine how many real solutions the equation has.

3x2 − 24x + 48 = 0

Solution

b2 − 4ac = (−24)2 − 4(3)(48) = 576− 576 = 0. Since the discriminate equals zero, the equationhas one (repeated) real solution.


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