A Real World Problem
An Application of Elementary Algebra
The obsession with “real world” problems in elementary mathematics courses has made it difficult for teachers to teach the fundamentals. Without mastery of those fundamentals, “real world” problems are impossible to solve. Consider the following problem. I will detail the background knowledge and skills necessary to solve this simple problem.
Problem: A rectangular garden plot has length 1 foot longer than twice the width. If the area of the garden is 55 square feet, what are the dimensions of the garden?
To begin solving this problem, students must understand what they are being asked to find. They must know that the dimensions of a rectangle are actually the length and width. Although that sounds simple, some students don’t make the connection, even though the words are contained in the first sentence.
Next, students must know how to calculate the area of a rectangle. That means they should know the formula for area of a rectangle; many donot. They use the formula for perimeter instead of area. Without this knowledge students cannot begin to solve this “real world” problem.
Now we move to the translation of the phrase “length 1 foot longer than twice the width”. The first step is to define a variable and clearly identify what the variable represents. Let W = the width of the rectangle. Then the length is L = 2W + 1. Since the area is 55 square feet, the equation to be solved is
LW = (2W + 1)W = 55
Even this step requires special knowledge; the variable L can be replaced by the expression 2W+1.
Studentsare now confronted with a major question:“What do I do with this equation?” They have just learned the zero-factor principle for solving quadratic equations by factoring. This equation does not look like the equations they solved successfully in the homework assignments. It does look like the equations they tried to solve, and quit trying to solve, when they didn’t get the answer in the back of the book.
To solve this equation, they need to remember that there should be a polynomial on one side of the equal sign and a 0 on the other side. If they remember that, the next challenge is to determine how to transform the equation
(2W + 1)W = 55
into
First, the parentheses must be removed. They need to apply the distributive property to transform the original equation into
Since a 0 should be on one side, the addition property of equality must be recalled and applied. Students will subtract 55 from both sides of the equation [no connection is made with the subtraction as the addition of the opposite]. The new equation will then be
OK. They can now apply the zero factor property. Well, not really;they don’t have any factors. There are two methods they have learned. One method reverses the FOIL process [Trial and Error] and the other, the ac-method, requires knowledge of factoring by grouping.
Both methods depend on the students’ knowledge of some elementary number facts. They must know basic multiplication and addition tables. They must be able to factor numbers into pairs of factors and then perform multiplication and addition to find the solution for this factoring problem. This process in itself can be time-consuming and discouraging if students have to perform “too many” calculations by pen & paper method. The cries of “Why can’t we use a calculator?” fill the air. The speed of mental calculations taken for granted by those of us who know our “tables” makes the educational process excruciatingly painful for students and teachers.
I will discuss the Trial and Error method.
Begin by noting that one of the factors must be (2W +__ ) since 2 is prime. First, students must know the difference between prime and composite numbers. They must then be able to identify numbers as either prime or composite. They also must know that 2W2 factors into 2WW so that the only possible factors begin with 2W and W.
What else must students know in order to use this method? Clearly they must be able to multiply polynomials. The multiplication process for this factoring method is usually FOIL. Most students are conditioned to use this process. Even so, they must also be able to identify and combine like terms (i.e. add polynomials).
Next, the factors of -55 must be grouped in pairs: {1, -55}, {-1,55}, {-5,11} and {5, -11}. Then the following pairs of binomial factors must be tested to find the correct product of factors.
a) (2W +1)(W - 55)b) (2W -55)(W + 1)
c) (2W - 1)(W +55)d) (2W +55)(W - 1)
e) (2W -5)(W +11)f) (2W +11)(W - 5)
g) (2W +5)(W -11)h) (2W -11)(W + 5)
The first option, a, is easily eliminated if the student quickly multiplies 2(-55). Sounds simple, but there is a negative number involved. Students break into cold (or hot) sweats and either quit trying to solve this problem, or perform the calculations incorrectly and use this factorization. [Note: This is a manual, not mental, calculation for students if calculators are not used.] If they use FOIL explicitly, this is the second multiplication (O = Outer terms). The second option, b, is also easily eliminated in the second multiplication.
In both cases, we assume that the student recalls the fact that the second term must equal W = 1W, the middle term of the quadratic polynomial we are trying to factor. Seems straightforward but students forget that the variable W has a coefficient of 1. Similar arguments eliminate c and d for students this is a time-consuming process).
The next factors require more work. In e, the student must again multiply and add positive and negative numbers. To perform the calculations correctly knowledge of the basic “tables” and operations with signed numbers is required. Otherwise the “security calculator” must save the day.Note: this solution is close to the correct factorization.Students may make an “error of convenience” because they are tired of working on this problem.
The next option, f, is similar to e. All student behavior stated previously applies, with one exception: if they multiply and add correctly, they will get the correct factorization.
(2W +11)(W - 5) = 2W 2 –10W +11W – 55 = 2W 2 +W – 55
Again, the negative numbers in the multiplication (2)(-5) and in the addition
–10 + 11 creates anxiety for many students.
Studentsare closer to a solution now that they have the correct factorization. The next step is to apply some more knowledge and skill recently acquired in the elementary algebra course. They must solve linear equations in one variable.These equations are the result of the zero-factor property, a property we hope that the student hasn’t forgotten at this point in the solution process. The property yields:
a) 2W +11 = 0orb) W – 5 = 0
The solution to b is clear. Students won’t use any solution method because they do remember that 5 - 5 = 0 so W = 5.
The solution of a will generate more anxious sweats. The student must apply the addition property of equality to get 2W = -11 and then apply the multiplication property of equality to get the solution W = -11/2, a fraction (!) and it has a negative sign (!!).
At this point, common sense kicks in (for most students). Students who have progressed this far usually reject the negative solution. With W = 5, L = 2(5) + 1 = 11. To find L the student used knowledge about evaluating algebraic expressions and about the order of operations in arithmetic.
In words, the dimensions of the rectangular garden are as follows: the length is 11 feet, and width is5 feet.
Hey! Not so fast. Did they check their answer? Since 5 feet 11 feet = 55 square feet, this really is the solution.
Is everyone as tired as I am? Imagine the students’ fear when trying to solve this elementary problem. These students are not children. They are adults, young and old, who hate (or at least strongly dislike) mathematics. For them, this is a “calculus problem”.
Joe Allen
4/14/2007