1. College Algebra Fifth Edition
James Stewart  Lothar Redlin  Saleem Watson

2. Equations and Inequalities1

3. Other Types of Equations
1.5

4. Other Types of Equations
So far, we learned how to solve linear and quadratic equations.
In this section, we study other types of equations, including those that involve:
Higher powers
Fractional expressions
Radicals

5. Polynomial Equations

6. Polynomial Equations
Some equations can be solved by factoring and using the Zero-Product Property.
The Zero-Product Property says that if a product equals 0, then at least one of the factors must equal 0.

7. E.g. 1—Solving an Equation by Factoring
Solve the equation
x5 = 9x3
We bring all terms to one side and then factor.

8. E.g. 1—Solving an Equation by Factoring
x5 = 9x3
x3(x2 – 9) = 0
x3(x – 3)(x + 3) = 0
x3 = 0 or x – 3 = 0 or x + 3 = 0
x = x = x = –3
You should check that each of these satisfies the original equation.

9. Caution
To divide each side of the equation in Example 1 by the common factor x3 would be wrong.
In doing so, we would lose the solution x = 0.
Never divide both sides of an equation by an expression that contains the variable unless you know that the expression cannot equal 0.

10. E.g. 2—Factoring by Grouping
Solve the equation
x3 + 3x2 – 4x – 12 = 0
The left-hand side of the equation can be factored by grouping the terms in pairs.

11. E.g. 2—Factoring by Grouping
(x3 + 3x2) – (4x + 12) = 0
x2(x + 3) – 4(x + 3) = 0
(x2 – 4)(x + 3) = 0
(x – 2)(x + 2)(x + 3) = 0
x – 2 = 0 or x + 2 = 0 or x + 3 = 0
x = x = – x = –3

12. E.g. 3—An Equation Involving Fractional Expressions
Solve the equation
To simplify the equation, we multiplying each side by the lowest common denominator (LCD).

13. E.g. 3—An Equation Involving Fractional Expressions

14. E.g. 3—An Equation Involving Fractional Expressions

15. E.g. 3—An Equation Involving Fractional Expressions
We must check our answers because multiplying by an expression that contains the variable can introduce extraneous solutions.

16. E.g. 3—An Equation Involving Fractional Expressions
We see that the solutions are x = 3 and –1.

17. Equations Involving Radicals

18. Equations Involving Radicals
When you solve an equation that involves radicals, you must be especially careful to check your final answers.
The next example demonstrates why.

19. E.g. 4—An Equation Involving a Radical
Solve the equation
To eliminate the square root, we first isolate it on one side of the equal sign, then square.

20. E.g. 4—An Equation Involving a Radical

21. E.g. 4—An Equation Involving a Radical
The values x = –¼ and x = 1 are only potential solutions.
We must check them to see if they satisfy the original equation.

22. E.g. 4—An Equation Involving a Radical
We see x = –¼ is a solution but x = 1 is not.

23. Extraneous Solutions
When we solve an equation, we may end up with one or more extraneous solutions—potential solutions that do not satisfy the original equation.
In Example 4, the value x = 1 is an extraneous solution.

24. Extraneous Solutions
Extraneous solutions may be introduced when we square each side of an equation because the operation of squaring can turn a false equation into a true one.
For example, –1 ≠ 1 but (–1)2 = 12

25. Extraneous Solutions
Thus, the squared equation may be true for more values of the variable than the original equation.
That is why you must always check your answers to make sure each satisfies the original equation.

26. Equations of Quadratic Type

27. Quadratic Type Equation
An equation of the form aW2 + bW + c = 0, where W is an algebraic expression, is an equation of quadratic type.
We solve these equations by substituting for the algebraic expression—as we see in the next two examples.

28. E.g. 5—An Equation of Quadratic Type
Find all solutions of:
We could solve this equation by multiplying it out first.
But it’s easier to think of the expression 1 + 1/x as the unknown and give it a new name W.

29. E.g. 5—An Equation of Quadratic Type
This turns the equation into a quadratic equation in the new variable W.

30. E.g. 5—An Equation of Quadratic Type
Now we change these values of W back into the corresponding values of x.

31. E.g. 6—A Fourth-Degree Equation of Quadratic Type
Find all solutions of:
x4 – 8x2 + 8 = 0
If we set W = x2, we get a quadratic equation in the new variable W.

32. E.g. 6—A Fourth-Degree Equation of Quadratic Type
(x2)2 – 8x2 + 8 = 0 (Write x4 as (x2)2)
W2 – 8W + 8 = 0 (Let W = x2)

33. E.g. 6—A Fourth-Degree Equation of Quadratic Type
So, there are four solutions:
Using a calculator, we obtain the approximations: x ≈ 2.61, 1.08, –2.61, –1.08

34. E.g. 7—An Equation Involving Fractional Powers
Find all solutions of:
x1/3 + x1/6 – 2 = 0
This equation is of quadratic type because, if we let W = x1/6, then W2 = (x1/6)2 = x1/3

35. E.g. 7—An Equation Involving Fractional Powers
x1/3 + x1/6 – 2 = 0
W2 + W – 2 = (Let W = x1/6)
(W – 1)(W + 2) = (Factor)
W – 1 = 0 or W + 2 = 0 (Zero-Product Property)

36. E.g. 7—An Equation Involving Fractional Powers
W = W = –2 (Solve)
x1/6 = x1/6 = –2 (W = x1/6)
x = 16 = x = (–2)6 = 64 (Take the 6th power)

37. E.g. 7—An Equation Involving Fractional Powers
By checking, we see that x = 1 is a solution but x = 64 is not.

38. Applications

39. Applications
Many real-life problems can be modeled with the types of equations that we have studied in this section.

40. E.g. 8—Dividing a Lottery Jackpot
A group of people come forward to claim a $1,000,000 lottery jackpot.
The winners are to share equally.

41. E.g. 8—Dividing a Lottery Jackpot
Before the jackpot is divided, three more winning ticket holders show up.
As a result, each person’s share is reduced by $75,000.
How many winners were in the original group?

42. E.g. 8—Dividing a Lottery Jackpot
We are asked for the number of people in the original group.
So, let: x = number of winners in the original group

43. E.g. 8—Dividing a Lottery Jackpot
We translate the information in the problem as follows.
In Words
In Algebra
Number of winners in original group x
Number of winners in final group
x + 3
Winnings per person, originally
Winnings per person, finally

44. E.g. 8—Dividing a Lottery Jackpot
Now, we set up the model.
Winnings per person, originally – 75, = Winnings per person, finally

45. E.g. 8—Dividing a Lottery Jackpot

46. E.g. 8—Dividing a Lottery Jackpot
Since we can’t have a negative number of people, we conclude that there were five winners in the original group.

47. E.g. 9—Energy Expended in Bird Flight
Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours.
Air generally rises over land and falls over water in the daytime.
So, flying over water requires more energy.

48. E.g. 9—Energy Expended in Bird Flight
A bird is released from point A on an island, 5 mi from B, the nearest point on a straight shoreline.
It flies to a point C on the shoreline.
Then, it flies along the shoreline to its nesting area D.

49. E.g. 9—Energy Expended in Bird Flight
Suppose it has 170 kcal of energy reserves.
It uses 10 kcal/mi flying over land.
It uses 14 kcal/mi flying over water.

50. E.g. 9—Energy Expended in Bird Flight
Where should C be located so that the bird uses exactly 170 kcal of energy during its flight?
Does it have enough energy reserves to fly directly from A to D?

51. E.g. 9—Energy in Bird Flight
Example (a)
We are asked to find the location of C.
So, let: x = distance from B to C

52. E.g. 9—Energy in Bird Flight
Example (a)
From the figure, and from the fact that energy used = energy per mile x miles flown we determine the following.

53. E.g. 9—Energy in Bird Flight
Example (a)
In Words
In Algebra
Distance from B to C x
Distance flown over water (from A to C)
Distance flown over land (from C to D)
12 – x
Energy used over water
Energy used over land
10(12 – x)

54. E.g. 9—Energy in Bird Flight
Example (a)
Now, we set up the model.
Total energy used = Energy used over water Energy used over land

55. E.g. 9—Energy in Bird Flight
Example (a)
Thus,
To solve this equation, we eliminate the square root by first bringing all other terms to the left of the equal sign and then squaring each side.

56. E.g. 9—Energy in Bird Flight
Example (a)

57. E.g. 9—Energy in Bird Flight
Example (a)
This equation could be factored.
However, as the numbers are so large, it is easier to use the quadratic formula and a calculator.

58. E.g. 9—Energy in Bird Flight
Example (a)
C should be either mi or mi from B so that the bird uses exactly 170 kcal of energy during its flight.

59. E.g. 9—Energy in Bird Flight
Example (b)
By the Pythagorean Theorem, the length of the route directly from A to D is:

60. E.g. 9—Energy in Bird Flight
Example (b)
Thus, the energy the bird requires for that route is: x 13 = 182 kcal
This is more energy than the bird has available.
So, it can’t use this route.