Domains of Rational Expressions

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close window McGraw-Hill McGraw-Hill Page 361 5.6 Solving Quadratic Equations by Factoring
    In This Section
    1 The Zero Factor Property
    2 Fractions and Decimals
    3 Applications

    The techniques of factoring can be used to solve equations involving polynomials. These equations cannot be solved by the other methods that you have learned. After you learn to solve equations by factoring,you will use this technique to solve some new types of problems.

    1 The Zero Factor Property

    In this chapter you learned to factor polynomials such as x2 + x − 6. The equation x2 + x − 6 = 0 is called a quadratic equation.

    Quadratic Equation

    If a, b, and c are real numbers with a ≠ 0, then

    is called a quadratic equation.

    A quadratic equation always has a second-degree term because it is specified in the definition that a is not zero. The main idea used to solve quadratic equations, the zero factor property, is simply a fact about multiplication by zero.

    The Zero Factor Property

    The equation a ⋅ b = 0 is equivalent to

    We will use the zero factor property most often to solve quadratic equations that have two factors, as shown in Example 1. However, this property holds for more than two factors as well. If a product of any number of factors is zero, then at least one of the factors is zero.

    The following strategy gives the steps to follow when solving a quadratic equation by factoring. Of course, this method applies only to quadratic equations in which the quadratic polynomial can be factored. Methods that can be used for solving all quadratic equations are presented in Chapter 10.

    Strategy for Solving an Equation by Factoring
    1. Rewrite the equation with 0 on one side.

    2. Factor the other side completely.

    3. Use the zero factor property to get simple linear equations.

    4. Solve the linear equations.

    5. Check the answer in the original equation.

    6. State the solution(s) to the original equation.

    Page 362 EXAMPLE 1 Using the zero factor property

    Solve x2 + x − 6 = 0.

    Solution

    First factor the polynomial on the left-hand side:

    We now check that −3 and 2 satisfy the original equation.

    The solutions to x2 + x − 6 = 0 are −3 and 2. Checking −3 and 2 in the factored form of the equation (x + 3)(x − 2) = 0 will help you understand the zero factor property:

    For each solution to the equation, one of the factors is zero and the other is not zero. All it takes to get a product of zero is one of the factors being zero.

    Now do Exercises 1–12

    Helpful Hint

    Some students grow up believing that the only way to solve an equation is to “do the same thing to each side.”Then along come quadratic equations and the zero factor prop erty. For a quadratic equation, we write an equivalent compound equation that is not obtained by“doing the same thing to each side.”

    A sentence such as x = −3 or x = 2, which is made up of two or more equations connected with the word “or,” is called a compound equation. In Example 2, we again solve a quadratic equation by using the zero factor property to write a compound equation.

    EXAMPLE 2 Using the zero factor property

    Solve the equation 3x2 = −3x.

    Solution

    First rewrite the equation with 0 on the right-hand side:

    Check 0 and −1 in the original equation 3x2 = −3x.

    There are two solutions to the original equation, 0 and −1.

    Now do Exercises 13–20

    Page 363 CAUTION

    If in Example 2 you divide each side of 3x2 = −3x by 3x, you would get x = −1 but not the solution x = 0. For this reason we usually do not divide each side of an equation by a variable.

    EXAMPLE 3 Using the zero factor property

    Solve (2x + 1)(x − 1) = 14.

    Solution

    To write the equation with 0 on the right-hand side, multiply the binomials on the left and then subtract 14 from each side:

    Check and 3 in the original equation:

    So the solutions are and 3.

    Now do Exercises 21–26

    CAUTION

    In Example 3, we started with a product equal to 14. Because 1 ⋅ 14 = 14, , and so on, we cannot make any conclusion about the factors that have a product of 14. If the product of two factors iszero, then we can conclude that one or the other factor is zero.

    If a perfect square trinomial occurs in a quadratic equation with 0 on one side, then there are two identical factors of the trinomial. In this case it is not necessary to set both factors equal to zero. The solution can be found from one factor.

    Page 364 EXAMPLE 4 An equation with a repeated factor

    Solve 5x2 − 30x + 45 = 0.

    Solution

    Notice that the trinomial on the left-hand side has a common factor:

    Even though x − 3 occurs twice as a factor, it is not necessary to write x − 3 = 0 or x − 3 = 0. If x = 3 in 5x2 − 30x + 45 = 0, we get

    which is correct. So the only solution to the equation is 3.

    Now do Exercises 27–30

    CAUTION

    Do not include 5 in the solution to Example 4. Dividing by 5 eliminates it. Instead of dividing by 5 we could have applied the zero factor property to 5(x − 3)2 = 0. Since 5 is not 0, we must have (x − 3)2 = 0 or x − 3 = 0.

    If the left-hand side of the equation has more than two factors, we can write an equivalent equation by setting each factor equal to zero.

    EXAMPLE 5 An equation with three solutions

    Solve 2x3 − x2 − 8x + 4 = 0.

    Solution

    We can factor the four-term polynomial by grouping:

    To check let x = − 2, , and 2 in 2x3 − x2 − 8x + 4 = 0:

    Since all of these equations are correct, the solutions are − 2, , and 2.

    Now do Exercises 31–38

    Helpful Hint

    Compare the number of solutions in Examples 1 through 5 to the degree of the poly nomial.The number of real solutions to any polynomial equation is less than or equal to the degree of the poly nomial. This fact is known as the fundamental theorem of algebra.

    Page 365 2 Fractions and Decimals

    If the coefficients in an equation are not integers, we might be able to convert them into integers. Fractions can be eliminated by multiplying each side of the equation by the least common denominator (LCD). To eliminate decimals multiply each side by the smallest power of 10 that will eliminate all of the decimals.

    EXAMPLE 6 Converting to Integers

    Solve.

    a)

    b)

    0.02x2 − 0.19x = 0.1

    Solution
    a)

    The LCD for 6 and 12 is 12. So multiply each side of the equation by 12:

    Check:

    The solutions are −6 and 4.

    b)

    Multiply each side by 100 to eliminate the decimals:

    The solutions are and 10. You might want to use a calculator to check.

    Now do Exercises 39–46

    CAUTION

    You can multiply each side of the equation in Example 6(a) by 12 to clear the fractions and get an equivalent equation, but multiplying the polynomial by 12 to clear the fractions is not allowed. It would result in an expression that is not equivalent to the original.

    Page 366

    Note that all of the equations in this section can be solved by factoring. However, we can have equations involving prime polynomials. Such equations cannot be solved by factoring but can be solved by the methods in Chapter 10.

    3 Applications

    There are many problems that can be solved by equations like those we have just discussed.

    Applications

    EXAMPLE 7 Area of a garden

    Merida's garden has a rectangular shape with a length that is 1 foot longer than twice the width. If the area of the garden is 55 square feet, then what are the dimensions of the garden?

    Solution

    If x represents the width of the garden, then 2x + 1 represents the length. See Fig. 5.1. Because the area of a rectangle is the length times the width, we can write the equation

    We must have zero on the right-hand side of the equation to use the zero factor property. So we rewrite the equation and then factor:

    The width is certainly not . So we use x = 5 to get the length:

    We check by multiplying 11 feet and 5 feet to get the area of 55 square feet. So the width is 5 ft, and the length is 11 ft.

    Figure 5.1

    Now do Exercises 65–66

    Helpful Hint

    To prove the Pythagorean theorem start with two identical squares with sides of length a + b, and partition them as shown.

    There are eight identical triangles in the diagram. Erasing four of them from each original square will leave smaller squares with areas a2, b2, and c2. Since the original squares had equal areas, the remaining areas must be equal.So a2 + b2 = c2.

    The Pythagorean theorem was one of the earliest theorems known to ancient civilizations. It is named for the Greek mathematician and philosopher Pythagoras. Builders from ancient to modern times have used the theorem to guarantee they had right angles when laying out foundations. The Pythagorean theorem says that in any right triangle the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

    The Pythagorean Theorem

    The triangle shown in Fig. 5.2 is a right triangle if and only if

    Figure 5.2 Page 367

    If you do an Internet search, you can find sites that have many different proofs to this theorem. One proof is shown in the Helpful Hint in the margin.

    EXAMPLE 8 Using the Pythagorean theorem

    The length of a rectangle is 1 meter longer than the width, and the diagonal measures 5 meters. What are the length and width?

    Solution

    If x represents the width of the rectangle, then x + 1 represents the length. Because the two sides are the legs of a right triangle, we can use the Pythagorean theorem to get a relationship between the length, width, and diagonal. See Fig. 5.3.

    To check this answer, we compute 32 + 42 = 52, or 9 + 16 = 25. So the rectangle is 3 meters by 4 meters.

    Figure 5.3

    Now do Exercises 67–68

    CAUTION

    The hypotenuse is the longest side of a right triangle. So if the lengths of the sides of a right triangle are 5 meters, 12 meters, and 13 meters, then the length of the hypotenuse is 13 meters, and 52 + 122 = 132.

    Warm-Ups Fill in the blank.
    1. A equation has the form ax2 + bx + c = 0 where a ≠ 0.

    2. A equation is two equations connected with the word “or.”

    3. The property says that if ab = 0, then a = 0 or b = 0.

    4. Some quadratic equations can be solved by .

    5. We do not usually each side of an equation bya variable.

    6. The theorem says that a triangle is a right triangle if and only if the sum of the squares of the legs is equal to the square of the hypotenuse.

    True or false?
    1. The equation x(x + 2) = 3 is equivalent to x = 3 or x + 2 = 3.

    2. Equations solved by factoring always have two different solutions.

    3. The equation ad = 0 is equivalent to a = 0 or d = 0.

    4. The solution set to (x − 1)(x + 4) = 0 is {1, −4}.

    5. If a, b, and c are the sides of any triangle, then a2 + b2 = c2.

    6. The solution set to 3(x − 4)(x − 5) = 0 is {3, 4, 5}.

    Page 368 Exercises Study Tips
    • Avoid cramming. When you have limited time to study for a test, start with class notes and homework assignments. Work one or two problems of each type.

    • Don't get discouraged if you cannot work the hardest problems. Instructors often ask some relatively easy questions to see if you understand the basics.

    1 The Zero Factor Property

    Solve by factoring. See Example 1. See the Strategy for Solving an Equation by Factoring box on page 361.

    1. (x + 5)(x + 4) = 0

    2. (a + 6)(a + 5) = 0

    3. x2 + 3x + 2 = 0

    4. x2 + 7x + 12 = 0

    5. w2 − 9w + 14 = 0

    6. t2 + 6t − 27 = 0

    7. y2 − 2y − 24 = 0

    8. q2 + 3q − 18 = 0

    Solve each equation. See Examples 2 and 3.

    1. x2 = x

    Exercise 14 - The Zero Factor Property: Already in ax2+bx+c = 0 Form

    • w2 = 2w

    • m2 = −7m

    • h2 = −5h

    • a2 + a = 20

    • p2 + p = 42

    • (x + 2)(x + 6) = 12

    Exercise 22 - The Zero Factor Property: Needs to be Put Into ax2+bx+c = 0 Form First

    • (x + 2)(x − 6) = 20

    Solve each equation. See Examples 4 and 5.

    1. 2x2 + 50 = 20x

    2. 3x2 + 48 = 24x

    Exercise 31 - The Zero Factor Property: Factor Out GCF With a Variable Then Factor Difference of Squares (3 solutions)

    • x3 − 9x = 0

    • 25x − x3 = 0

    • w3 + 4w2 − 4w = 16

    • a3 + 2a2 − a = 2

    • n3 − 3n2 + 3 = n

    • w3 + w2 − 25w = 25

    2 Fractions and Decimals

    Solve each equation. See Example 6.

    1. 0.01x2 + 0.08x = 0.2

    2. 0.01x2 − 0.07x = −0.1

    3. 0.1x2 + 0.7x + 1 = 0

    Page 369 Miscellaneous

    Solve each equation.

    1. x2 − 16 = 0

    2. x2 − 36 = 0

    3. a3 = a

    4. x3 = 4

    5. 3x2 + 15x + 18 = 0

    6. −2x2 − 2x + 24 = 0

    7. (t − 3)(t + 5) = 9

    8. (x − 2)2 + x2 = 10

    9. (x − 3)2 + (x + 2)2 = 17

    Exercise 62 - Solve the equation

    • a3 + 3a2 − 25a = 75

    • m4 + m3 = 100m2 + 100m

    3 Applications

    Solve each problem. See Examples 7 and 8.

    1. Dimensions of a rectangle. The perimeter of a rectangle is 34 feet, and the diagonal is 13 feet long. What are the length and width of the rectangle?

    2. Address book. The perimeter of the cover of an address book is 14 inches, and the diagonal measures 5 inches. What are the length and width of the cover?

      Figure for Exercise 66
    3. Violla's bathroom. The length of Violla's bathroom is 2 feet longer than twice the width. If the diagonal measures 13 feet, then what are the length and width?

    Exercise 68 - Applications: Rectangular Stage Dimensions

    • Rectangular stage. One side of a rectangular stage is 2 meters longer than the other. If the diagonal is 10 meters, then what are the lengths of the sides?

      Figure for Exercise 68
    • Consecutive integers. The sum of the squares of two consecutive integers is 13. Find the integers.

    • Consecutive integers. The sum of the squares of two consecutive even integers is 52. Find the integers.

    • Two numbers. The sum of two numbers is 11, and their product is 30. Find the numbers.

    • Missing ages. Molly's age is twice Anita's. If the sum of the squares of their ages is 80, then what are their ages?

    • Three even integers. The sum of the squares of three consecutive even integers is 116. Find the integers.

    • Two odd integers. The product of two consecutive odd integers is 63. Find the integers.

    • Consecutive integers. The product of two consecutive integers is 5 more than their sum. Find the integers.

    • Consecutive even integers. If the product of two consecutive even integers is 34 larger than their sum, then what are the integers?

    • Two integers. Two integers differ by 5. If the sum of their squares is 53, then what are the integers?

    • Two negative integers. Two negative integers have a sum of −10. If the sum of their squares is 68, then what are the integers?

    • Lucy's kids. The sum of the squares of the ages of Lucy's two kids is 100. If the boy is two years older than the girl, then what are their ages?

    Page 370
    • Sheri's kids. The sum of the squares of the ages of Sheri's three kids is 114. If the twin girls are three years younger than the boy, then what are their ages?

    • Area of a rectangle. The area of a rectangle is 72 square feet. If the length is 6 feet longer than the width, then what are the length and the width?

    • Area of a triangle. The base of a triangle is 4 inches longer than the height. If its area is 70 square inches, then what are the base and the height?

    • Legs of a right triangle. The hypotenuse of a right triangle is 15 meters. If one leg is 3 meters longer than the other, then what are the lengths of the legs?

    • Legs of a right triangle. If the longer leg of a right triangle is 1 cm longer than the shorter leg and the hypotenuse is 5 cm, then what are the lengths of the legs?

    • Skydiving. If there were no air resistance, then the height (in feet) above the earth for a skydiver t seconds after jumping from an airplane at 10,000 feet would be given by

      a)

      Find the time that it would take to fall to earth with no air resistance; that is, find t for which h(t) = 0. A skydiver actually gets about twice as much free fall time due to air resistance.

      b)

      Use the accompanying graph to determine whether the skydiver (with no air resistance) falls farther in the first 5 seconds or the last 5 seconds of the fall.

      c)

      Is the skydiver's velocity increasing or decreasing as she falls?

      Figure for Exercise 85
    • Skydiving. If a skydiver jumps from an airplane at a height of 8256 feet, then for the first five seconds, her height above the earth is approximated by the formula h(t) = −16t2 + 8256. How many seconds does it take her to reach 8000 feet?

    • Throwing a sandbag. A balloonist throws a sandbag downward at 24 feet per second from an altitude of 720 feet. Its height (in feet) above the ground after t seconds is given by S(t) = −16t2 − 24t + 720.

      a)

      Find S(1).

      b)

      What is the height of the sandbag 2 seconds after it is thrown?

      c)

      How long does it take for the sandbag to reach the ground? [On the ground, S(t) = 0.]

    • Throwing a wrench. An angry construction worker throws his wrench downward from a height of 128 feet with an initial velocity of 32 feet per second. The height of the wrench above the ground after t seconds is given by S(t) = −16t2 − 32t + 128.

      a)

      What is the height of the wrench after 1 second?

      b)

      How long does it take for the wrench to reach the ground?

    • Glass prism. One end of a glass prism is in the shape of a triangle with a height that is 1 inch longer than twice the base. If the area of the triangle is 39 square inches, then how long are the base and height?

      Figure for Exercise 89
    • Areas of two circles. The radius of a circle is 1 meter longer than the radius of another circle. If their areas differ by 5 pi; square meters, then what is the radius of each?

    • Changing area. Last year Otto's garden was square. This year he plans to make it smaller by shortening one side 5 feet and the other 8 feet. If the area of the smaller garden will be 180 square feet, then what was the size of Otto's garden last year?

    • Dimensions of a box. Rosita's Christmas present from Carlos is in a box that has a width that is 3 inches shorter than the height. The length of the base is 5 inches longer than the height. If the area of the base is 84 square inches, then what is the height of the package?

      Figure for Exercise 92
    Page 371
    • Flying a kite. Imelda and Gordon have designed a new kite. While Imelda is flying the kite, Gordon is standing directly below it. The kite is designed so that its altitude is always 20 feet larger than the distance between Imelda and Gordon. What is the altitude of the kite when it is 100 feet from Imelda?

    • Avoiding a collision. A car is traveling on a road that is perpendicular to a railroad track. When the car is 30 meters from the crossing, the car's new collision detector warns the driver that there is a train 50 meters from the car and heading toward the same crossing. How far is the train from the crossing?

    • Carpeting two rooms. Virginia is buying carpet for two square rooms. One room is 3 yards wider than the other. If she needs 45 square yards of carpet, then what are the dimensions of each room?

    • Winter wheat. While finding the amount of seed neededto plant his three square wheat fields, Hank observed that the side of one field was 1 kilometer longer than the sideof the smallest field and that the side of the largest field was 3 kilometers longer than the side of the smallest field. If the total area of the three fields is 38 square kilometers, then what is the area of each field?

    • Sailing to Miami. At point A the captain of a ship determined that the distance to Miami was 13 miles. If she sailed north to point B and then west to Miami,the distance would be 17 miles. If the distance from point A to point B is greater than the distance frompoint B to Miami, then how far is it from point A to point B?

      Figure for Exercise 97
    • Buried treasure. Ahmed has half of a treasure map,which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x + 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x?

    • Broken Bamboo I. A 10 chi high bamboo stalk is broken by the wind. The top touches the ground 3 chi from its base as shown in the accompanying figure. At what height did the stalk break? This problem appeared in a book by Chinese mathematician Yang Hui in 1261.

      Figure for Exercise 99
    • Broken Bamboo II. A section of bamboo that is 5 chi in length is broken from a stalk of bamboo of unknown height. If the broken section touches the ground 3 chi from the base as in Exercise 99, then what was the original height of the bamboo stalk?

    • Emerging markets. Catarina's investment of $16,000 in an emerging market fund grew to $25,000 in two years. Find the average annual rate of return by solving the equation 16,000(1 + r)2 = 25,000.

    • Venture capital. Henry invested $12,000 in a new restaurant. When the restaurant was sold two years later, he received $27,000. Find his average annual return by solving the equation 12,000(1 + r)2 = 27,000.

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