Chapter 1 — Equations and Inequalities

Transcription

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Chapter 1 is partly a review of algebra you have learned before, and partly newer material. If you haven’t
done algebra for a while then this could be a challenging chapter for you and you will need to give it some extra
time and effort.
Each section is divided into topics. At the end of most topics are one or two problems called “reinforcement
problems.” You should work these, at a minimum, before reading the next topic. Solutions to the reinforcement
problems are worked out at the end of each section.
Each section introduction also lists the problems from the course textbook that I recommend you practice.
Bold print type, when it appears in a paragraph, indicates a new term that is being introduced.
Section 1.1 — Linear Equations
This section is a review of solving linear equations and rational equations (equations with fractions in them).
College Algebra, as you might imagine, steps up the difficulty a bit. I want to draw your attention more to the
process of solving. A clear understanding of the solving process will serve you well in this course.
Topic 1:
Topic 2:
Topic 3:
Topic 4:
Topic 5:
Topic 6:
Topic 7:
Topic 8:
Solving linear equations.
An unsolvable linear equation.
The geometry behind linear equations.
Solving a simple rational equation.
Solving a more complicated rational equation.
Division by zero.
A rational equation with no solution.
Solving literal equations.
Practice Problems: 17 – 81 odd, skip 75.
Topic 1: Solving linear equations.
Solve:
Linear equations are first degree polynomial equations. Although fractions can appear, the variable is not
part of a denominator. All linear equations, ultimately, can be reduced to the simple form ax + b = 0. First
degree polynomials contain variables that have 1 as the largest exponent. The term, “linear,” from the root word
“line,” derives from the fact that a linear equation is the representation of a line. I will demonstrate this later
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and show why linear equations have at most one solution.
I’m assuming that you have some experience at solving linear equations, so I have led off with a more
complicated problem than you may be accustomed to in order to give a full review of the process of solving
such equations. Now that you have reached the level of College Algebra you should know that the emphasis is
now shifting from learning techniques of algebra and more towards analysis of mathematics using algebra. The
more you focus on meaning the easier it will be for you to do well.
What is the first thing we should do with the problem above? Although there are good exceptions to some
rules, the reliable first step in solving any equation is to simplify each side separately. In this equation the right
side contains a parentheses inside a set of brackets. Brackets function identically to parentheses, but are normally
used to enclose parentheses. We use brackets and parentheses both in order to make equations more readable.
When something is needed to enclose brackets, then we use braces: { }. All three symbols, parentheses, brackets
and braces, are called grouping symbols, literally symbols that group terms together.
It is a good policy to simplify expressions from the inside out, so on the right hand side I will first remove
the parentheses inside the brackets.
Removing the parentheses on both side of the equation used the distributive law. The distributive law is the
axiom of the real numbers which combines multiplication/division with addition/subtraction. People who are
rusty at algebra often carelessly make errors when applying the distributive law. I will show the distributive
step on both sides of the equation this way.
Next we can combine like terms on both sides of the equation in order to shorten the equation.
Now we can remove the brackets by distributing the 2, and then finish simplifying the right side.
All linear equations can be simplified to this point, where there are at most four terms. Now we need to
move things about to finish solving. Our goal is to isolate x. We can only change an equation by performing
the same operation on both sides, such as adding 3x to both sides.
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We continue trying to isolate x, this time eliminating the – 3 on the right side by adding 3 to both sides. Note
that in this familiar process of solving we are eliminating terms or numbers by a reverse process. By this I mean
that 3 is currently being subtracted from the right side. In order to eliminate this, we do the opposite, add 3.
Finally, what is 25 doing to x? Since it is multiplying x, we eliminate the 25 from the right side by doing the
opposite, dividing by 25.
The sensible thing to do is to write the answer with x on the left side. In English we would say, “x is 13/25,”
not “13/25 is x.”
Finally, how do you know it’s right? A solution to an equation is one which balances both sides. I will
normally leave the checking to you, but I’ll show this one as a review of handling fractions.
Of course, if the left side does not equal the right side then there is a mistake, either in the solution or in
the checking.
In a later chapter I will return to problems like these and examine the solving process in the context of
inverse functions. Meanwhile, it is a great idea to work the reinforcement problems presented at the end of most
topics while you are fresh from reading, before moving on to the next topic. Don’t assume that if you understand
what you just read then you don’t need to practice! Solutions are given at the end of each section.
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Reinforcement Problems
1.
Solve:
2.
Solve:
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Topic 2: An unsolvable linear equation.
You will occasionally encounter oddities in this course. No one is trying to trick you, but being alert for
exceptions becomes important as you work your way up the math food chain.
Our first oddity is quite simple.
Solve: x = x + 1
What can be done? If you subtract x from both sides, which is the only way to make progress, the result
is 0 = 1. This implies that no matter what number x could be, we always end up with the impossible 0 = 1.
It’s important that you communicate your conclusion to every problem.
How does this happen? In this example the impossibility should be clear. If you translate the problem
into English it would say that here is a number equal to one more than itself. Obviously, no number works that
way.
We will occasionally run into problems which can’t be solved in this course but it will not always be obvious
that the problem has no solution. It is possible to follow all the rules of solving correctly and yet still produce
a false answer. You will see the first example of this in Topic 6.
I want to leave you with this important distinction because so many students get this confused.
When an impossible situation such as 2 = 8 arises as a result of correctly solving an
equation, the equation has no solution.
When an impossible situation such as 2 = 8 arises as a result of checking the answer
to an equation, then either the solution is wrong and needs to be fixed, or there is an
error in the arithmetic of the check.
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Topic 3: The geometry behind linear equations.
Arithmetic and geometry are the two founding branches of mathematics, going back thousands of years.
Arithmetic arose out of a need to count, and geometry arose out of a need to understand our physical world.
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Although people did dabble at algebra in very primitive ways, algebra only began to become established around
825 a.d., in Persia. Algebra as we would commonly recognize it was an innovation of a Frenchman, François
Viète, (1540 – 1603), in 1591. Viète created the system of arithmetic symbols and the use of letters to represent
variables that we now take for granted. Soon thereafter, another Frenchman, René Descartes (1596 – 1650)
discovered how geometry and algebra were strongly linked. Among his many achievements, Descartes was the
one who came up with the x–y plane and graphing. For the first time, someone had linked the powerful analytical tools of algebra with geometry. He called this analytical geometry, a title which still appears in some
calculus books.
As it turns out, then, algebra and geometry share a great deal, so much so that it is possible to work some
geometry problems by using algebra, and to work some algebra problems by using geometry. In fact, this crossover will explain a lot of what you will see in this course.
Solving linear equations is the equivalent of finding the intersection of lines. I’ll give a simple example
in order illustrate this.
Solve: 2x – 3 = – x + 3
I’m sure that you will have no difficulty in solving this little equation and getting the answer, x = 2. What
gets interesting is if you let each side of the equation equal y and graph each side as a line, as follows.
y = –x + 3
y = 2x – 3
2
1
1
2
As you can see, the intersection point of the lines is the point (2, 1), in which x = 2.
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The two sides of any linear equation represent two lines on a graph, and the x coordinate of the intersection
point is the solution to the equation. Since two different lines cannot cross more than once, this demonstrates
that there cannot be more than one solution to a linear equation.
What if the lines are parallel? Parallel lines will not intersect, so there is no solution to the equation since,
again, the intersection point is the geometrical representation of the algebra solution. This was the case with
the equation we saw earlier that could not be solved: x = x + 1. Both lines, y = x, and y = x + 1, have the
same slope but different y–intercepts, so they run in the same direction—parallel—but do not intersect.
The weak link in all this discussion is the assertion that linear equations actually represent lines. We will
prove that to be true in the next chapter.
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Topic 4: Solving a simple rational equation.
Solve:
This simple, rational equation allows us to illustrate a reliable way to eliminate the complication of fractions in equations. The word “ratio” is the root word of “rational.” In the context of mathematics a rational
expression, function or equation is one containing at least one fraction. A rational number is one that can be
expressed as a ratio (fraction) of integers.
The least common denominator (LCD) of two or more fractions is the smallest term into which each denominator will divide evenly. In this problem, the LCD is 2x. We can use the LCD to eliminate the fractions.
In effect, by multiplying every term of the equation by the LCD we are rescaling the equation in a way that
eliminates the fractions. Notice that each term of the equation must be multiplied by the least common denominator, including the non-fraction, 2.
As long as you have used a legitimate common denominator, then each denominator will cancel off. From
there we have an ordinary equation to solve.
The remaining steps, hopefully, are routine for you.
It is important to clearly write your final answer.
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As we shall see shortly, it is a really good idea to check your answers when solving rational equations.
Reinforcement Problem
3.
Solve:
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Topic 5: Solving a more complicated rational equation.
Solve:
This rational equation clearly requires more work than the previous example, and thus has more pitfalls.
When I mention parts of problems where students make common mistakes you should pay close attention. These
are the mistakes you are most likely to make!
Let’s begin by factoring the denominator(s) when possible, and organizing things by enclosing binomial
terms with parentheses. The more organized you are the less chances there are for careless errors. Notice also
how I’ve spread out the terms so that we can write things in without too much clutter. Avoid cramping your
work because people make mistakes doing that. Paper is cheap, use it!
The least common denominator is the product of the highest powers of the different factors. In this case,
the LCD is (x – 1)(x + 1). As before, multiplying by the LCD cancels each denominator.
This simplifies nicely and leaves us a simple linear equation to solve.
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I think it a very good idea to write down what is left before finishing the problem. Students who skip this
step make a distribution error the majority of the time.
The common error in this problem is in getting 3x – 2x + 2, or 3x – 2x + 1, on the left side. Students who are
too lazy to write parentheses almost always make this type of distribution error, but those who do use parentheses
but skip this step very often make this careless mistake also. Good mathematics requires great organization both
in the written form and in the mind while solving. If you cut corners then don’t expect to do very well.
As mentioned earlier, it is an especially good idea to check your answers when solving rational equations.
This one will be a little messy, but it’s only arithmetic.
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4.
Solve:
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Topic 6: Division by zero.
Simplify:
It is important that we review the issue of division by zero because it can arise whenever we are solving
rational equations. The naive answer to 4 ÷ 0 is 0. The stock answer textbooks give is “undefined.” What
does all this mean?
How did you learn how to divide? Most people learned division of small numbers as an inverse of multiplication. For example, we know that 24 ÷ 8 = 3 because 8 • 3 = 24. I’ll write that out so that it is more
visual.
So, let’s try the same with 4 ÷ 0 = 0.
In fact, no matter what number you put on the right side, 4 ÷ 0 cannot equal it since 0 • number = 0. So
we need to conclude that 4 ÷ 0 cannot be done. When textbooks say, “undefined,” what they really mean is
that division by zero is an undefined operation, one for which there is no possible answer. In plainer English,
which I tend to prefer, you might conclude the following.
For that matter, division of anything by zero is impossible, or cannot be done. Or is it? Consider the following division by zero.
Simplify:
Why couldn’t 0 ÷ 0 = 0? We agree that 0 • 0 = 0, so the division checks. Perhaps this is an exception
to the rule? You see, if we agree that 0 ÷ 0 = 0 then we also must agree that 0 ÷ 0 = 17. Seventeen fits as
an answer, right? 0 • 17 = 0. So does any number. It appears that any number could represent 0 ÷ 0.
There are several strange forms of arithmetic computations such as 0 ÷ 0, called indeterminate forms.
This particular indeterminate form lies at the heart of differential calculus. It is not that we decide what 0 ÷ 0
equals through calculus. Instead, we are concerned about what happens when you get immeasurably close to
the 0 ÷ 0 situation. Unfortunately, we’ll need to leave that fascinating issue to a calculus course.
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Topic 7: A rational equation with no solution.
Solve:
As before, let’s solve this by first clearing the fractions by multiplying both sides by the least common
denominator, x – 5. Again, each term is multiplied, even the non-fraction, 2.
Cancelling like factors leaves a routine equation, which I’ll proceed to solve.
Checking this answer gives us an interesting result.
Our purported answer leads to an impossible calculation, 5 ÷ 0. What now? Provided that we made no mistakes
in solving, there is no number which can balance this equation, so our conclusion can only be the following.
But haven’t we made an error somewhere when solving? We did get an answer, x = 5, but it didn’t work.
How can that be? I’ll let you ponder that while we take a different approach to solving this problem.
Solve:
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Few rational equations lend themselves easily to the method I am about to use because most do not come
with common denominators. In this case we can subtract a term on the right in order to rearrange the equation
to our convenience.
But now we can easily combine the two terms on the left into a single fraction since they have common denominators.
Of course, and this is the punch line of this problem, the left side reduces to 1, producing an impossible situation
regardless of what number we let x equal.
This again leaves us with only one conclusion about this problem.
The nice thing about the second method of solving this particular problem is that it shows that there is no
possible way to solve it. So how did we get the false (extraneous) answer of x = 5 in the first method? It’s an
important question because resolving this conundrum will shed light on how we can go astray.
Let’s revisit our strategy at the beginning of this topic, multiplying the equation by the common denominator.
Without knowing it, since we ended up with x = 5, by what number did we multiply both sides of the equation?
Since x – 5 = 5 – 5 = 0, we multiplied the equation by 0. Is that okay? Of course not! Everyone learns
that it’s okay to multiply an equation by any number they want to, but many don’t learn, or forget, that there is
an exception to this rule.
Let’s start with an absurdity.
Now let’s multiply the equation by 0 and see what happens.
So multiplying this false equation by zero turned it into a true equation. In a sense, that’s what happened at the
beginning of this topic when we multiplied by x – 5, unwittingly multiplying an unsolvable equation by zero,
thus turning it into something that could be solved.
The bad news is that we cannot predict in advance when we are accidentally multiplying a false equation
by zero. Our only recourse is to check each answer. Some students misunderstand the results of checking equa-
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tions, however. Here is an observation similar to one I made in Topic 2 that would be well to heed.
If the solution to an equation causes division by zero when substituted back into the
equation, then that is not a solution.
If the solution to an equation causes the left side to not equal to the right side, then either
your solution is wrong, or there is a mistake in the check.
5.
Solve:
6.
Solve:
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Topic 8: Solving literal equations.
Solve for
:
This is a rather common formula from statistics, comparing the mean of a sample, , to a set value, µ.
These kinds of equations, which contain more than one letter, are sometimes called literal equations. When
solving literal equations we are solving for a specified variable while treating the other letters as if they were
constants, i.e., numbers that do not change. Thus a literal equation has only one variable in it, the one specified
that we are to solve.
As you become more advanced in your mathematics learning you are called upon to work in more generalities than specifics, such as equations with letters representing constants rather than specific equations with
given numbers as constants.
There is no secret to solving literal equations other than viewing all the letters as if they were numbers,
except for the letter for which you are solving. In this case we want to isolate , so we can begin by eliminating
the fraction from the side with .
The next step eliminates µ just as you would any negative number from one side of an equation.
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It would be conventional to rewrite this with the solved variable on the left. Otherwise, it is as if you are
talking backwards.
________________________________________
Solve for p:
This formula comes from the study of optics. The f-stop setting on a 35 mm camera is the same as letter
f in this equation.
As with any rational equation, the best place to start is with eliminating the denominators by multiplying
all terms by the least common denominator.
After cancelling like factors in order to simplify, we have the following.
Students unfamiliar with solving this type of equation are often stumped at this point. How do you think
through these situations? Since we are solving for p, shouldn’t we bring terms with p together on one side and
keep the other term by itself?
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Notice that each term on the left has a p in it. We can further the isolation of p by factoring it out.
The finish may now be obvious. We divide both sides by the term that does not contain p: (q – f).
7.
Solve for v0:
8.
Solve for y´:
from calculus.)
(Note: y´ is not the same as y1. This problem comes
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Solutions to Reinforcement Problems
(Note: I have not shown the checks, but that doesn’t mean you should not check these answers yourself.
I did!)
1.
2.
3.
4.
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5.
6.
7.
8.
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Section 1.2 — Quadratic Equations
This section is a review of solving quadratic equations, but also includes some applications (word problems).
After linear equations, quadratic equations are quite common in applications.
Topic 1:
Topic 2:
Topic 3:
Topic 4:
Topic 5:
Topic 6:
Topic 7:
Topic 8:
Solving equations by factoring.
Solving by using the square root.
Solving by completing the square.
The quadratic formula.
Applying the quadratic formula.
Calculating approximate solutions.
A basic quadratic application.
Another quadratic application.
Practice Problems: 9 – 33, 47 – 91, 99 – 119, all odd.
Topic 1: Solving equations by factoring.
Solve:
A quadratic equation is a second degree polynomial equation. By second degree we mean that the largest
exponent of the polynomial is 2. While quadratics are the topic of this section, the method of factoring works
on any equation that can be factored, and indeed, the fact that we can use factoring to solve equations is a major
reason for learning factoring in algebra.
The method of factoring only works when all the terms are on one side of the equation and zero is on the
other. I’ll explain that shortly. This problem is ready to factor.
An important comment is in order here. About the only way students get these problems wrong is because
they do not check their factoring. To check any factoring you multiply everything back to see if it gets you back
to the original equation. After 35 years of doing this stuff I still check my factoring!
Now what? One of the hidden beauties of the real numbers is that whenever a product of numbers gives
you 0, then one of the numbers must be 0. Have you seen it otherwise? This property of the real numbers is
known as the zero product property. There are algebraic systems without this property, but not in the ordinary
real number system. So the only possible solution to this problem is one of two possibilities, either the first
factor, x – 3, is zero, or the second factor, x – 5, is zero. Both of these little equations can be solved, so our
problem has two answers.
Therefore, the next step in solving is letting each factor equal zero.
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This gives us the two answers.
Getting two answers for a quadratic is usual, although we will see that sometimes both answers must
be expressed with a new type of number, complex numbers. We will confront those in the next section. The
reason for two answers, of course, is related to the degree of the polynomial, 2. Geometrically, if we represent
the left side of the equation in the above problem as
, the graph is of an object known as
a parabola. This curved object crosses the x–axis in two places, x = 3 and x = 5, as shown below. (Each
scale mark is one unit.)
Later in this course we will analyze the graphs of quadratics and other polynomials. Higher degree polynomial
equations that can be factored can also be solved by employing the zero product property, and later we’ll also
study ways to factor those higher degree equations.
________________________________________
Solve:
While there is a zero product property, there is no “six product property.” In other words, it makes no
sense to factor the left side as is because there are infinitely many ways to multiply two numbers to get 6. The
sensible procedure is to subtract the 6 from both sides and proceed to solve by factoring by using the zero product property.
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Solve:
This is not a true quadratic equation, but it will become one once we clear the fractions. As before, we
multiply each term by the common denominator in order to eliminate the denominators.
Note the organized use of parentheses. Don’t try to skip too many steps; that could be careless.
Now we rearrange the terms by subtracting 2x + 2 from both sides, leaving a quadratic equation that can be
solved by factoring.
Since this was a rational equation to start with, it is a good idea to at least confirm that neither answer causes
division by zero when substituted back into the original equation.
1.
Solve:
2.
Solve:
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Topic 2: Solving by using the square root.
Solve:
At a glance it should be obvious that x = 5 is the solution. It is less obvious that x = – 5 is also a solution. A methodical way to solve the equation and get both solutions is to use factoring.
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Our problem above, when written as
, involves factoring a difference of two squares. We
can, in fact, generalize all difference of squares equations like this.
When factored and solved we get two answers which differ only in sign.
Because this happens every time, we can develop an alternative technique of solving by using the square
root. Returning to the original problem,
, we see that taking the square root of each side will give us
x = 5, but since there will always be two answers differing only in sign, then we need to adjust this technique
to include both the positive and negative answers by using the plus or minus sign: ± .
Of course, this still means we have two separate answers.
Let’s summarize this square root technique.
When solving an equation by taking square roots of both sides, use the ± symbol on one
side in order to account for both possible answers.
Another way of looking at this, technically, is that the square root of a variable term produces an absolute
value; e.g.,
. In that case
, an equation with two answers, 5 and – 5. We use the absolute
value in this case because we define the square root of a positive number as a positive number.
________________________________________
Solve:
We now begin to see the usefulness of this square root technique of solving. Were you to multiply this
equation out and combine terms you would get
, an equation that cannot be factored
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in any conventional sense. We can still solve this equation by first isolating the constant term.
The equation is now in the proper form for taking the square root of each side. Just to emphasize an important
feature of this method, there cannot be any variables on one side of the equation; i.e., the constant 7 is isolated
from the square of the variable term.
Note again how the use of the “±” symbol will give us two answers to this equation. The rest is routine solving.
3.
Solve:
4.
Solve:
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Topic 3: Solving by completing the square.
As a prelude to learning the quadratic formula I always like to walk students through a process of solving
quadratic equations known as completing the square. It takes its name because it is a method of manipulating
a quadratic equation into the form that can be solved by the square root method outlined above. To make this a
little more interesting, I’ll illustrate this on a geometric form known in ancient times as the “golden rectangle.”
Here is how such a rectangle is defined.
The golden rectangle is one such that if the square formed by the short side is removed,
then the remaining rectangle has the same proportions as the original.
On the next page is a scale rendering of a golden rectangle, and I have taken the liberty of labeling the sides,
with the original short side measuring one unit, as well as drawing the square that can be removed. Length x
represents the length of the long side of the golden rectangle. When the shorter side is length 1, then the number
x is known as the golden mean.
As an aside, ancient Greeks and Romans considered the golden rectangle the most aesthetically pleasing
rectangle. Rectangles with this proportion appear in much ancient architecture from those cultures, and had a
resurgence in art in the middle ages. Even Leonardo da Vinci used golden rectangles in painting the Mona Lisa.
For example, her head can be enclosed within such a rectangle. In the 20th century, artist Piet Mondrian made
a good living drawing rectangles, often deliberately using the golden rectangle in his abstract art.
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The golden mean surfaces in many areas of nature. Entire books have been written about this.
1
1
1
x
x – 1
Returning to our problem, we must use the definition of a golden rectangle in order to set up an equation
that can be solved. Recall that the proportional relationship between sides is maintained after the square is removed. Let’s compare the ratios of the long side to the short side.
Now we have an equation to solve. We’ll clear the fractions in the usual manner, multiplying by the common
denominator.
We can finish rearranging this equation by subtracting 1 from each side to produce
,
which is a bona fide quadratic equation. It is not possible to solve this by factoring in the traditional sense, but
it turns out that the first step in solving this equation by completing the square is to send the 1 back to the right
side anyway.
A big note is in order here. The equation as we left it above is not in the form which lends itself to the
square root method. We have isolated a constant, 1, but the left side is not a perfect square. Taking the square
root of
is unproductive since
cannot be simplified.
The next step in completing the square is by no means obvious. We take half of the coefficient of the x
term, half of – 1 that is, square this number, and add it to both sides of the equation.
It turns out that every time we follow this process, provided that the leading coefficient, i.e., the coefficient
of x , is 1, then the resulting trinomial can be factored.
2
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Matter of fact, not only can the trinomial be factored, it can be factored as a perfect square.
Now the equation has been manipulated into the form that we can now solve by using the square root
method. That is the purpose of completing the square.
It appears that we can finish by adding the half to each side.
It is conventional, however, to combine the fractions, as follows.
Thus completes the square. Since this “word problem” can only have one answer, we need to dissect the
result above and determine the correct golden mean. You should try to calculate these on your calculator to see
if you get the same result. If you have trouble getting the correct numbers, Topic 6 is devoted to computing
solutions to quadratic equations.
First we separate the two answers.
Now we calculate each answer separately. When entering these into your calculator, you’ll want to put each
numerator in parentheses so that your calculator knows to complete the numerator calculation before dividing
by the 2.
I have rounded these off to three decimal places, which is plenty of precision in many circumstances.
The answer to our problem can only be the positive result, 1.618, since we cannot have a side of a rectangle
with negative length. Actually, when quadratic equations surface in applications it is not uncommon to get two
answers with one answer correct and the other impossible. Such impossible answers are known as extraneous
solutions. These bogus solutions would exist in a world where negative lengths are possible.
Here is the golden mean calculated to all the decimals my calculator will allow.
As mentioned, this is an interesting number due to the many ways in which it shows up in nature and in
geometry. It is interesting for two additional features. Squaring it is the same as adding 1. Taking the reciprocal is the same as subtracting 1.
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Topic 4: The quadratic formula.
All College Algebra students should study the derivation of the quadratic formula. The quadratic formula is
the solution to the general quadratic equation,
. As a more abstract equation using symbols
instead of specific constants (numbers), the concentration you place in trying to understand this derivation will
help you to adjust to working with other abstract looking equations and problems. Furthermore, the derivation
of the quadratic formula was a seminal moment in algebra: the solution to an entire class of equations. In other
words, we will now proceed to solve every possible quadratic equation.
Every quadratic equation can be manipulated into the general form
. Our solution
is actually an outline of the completing the square method for solving quadratics.
First we want to create a quadratic polynomial in which the leading coefficient is 1. We do this by dividing all terms by a.
An important consideration to the above step is that we not divide by 0. But
would not
be a quadratic equation if a = 0, would it? In that case it would be a linear equation, which would be solved
by simpler means.
Our next step is to isolate the constant term c / a.
Next we use that unusual step seen in the previous topic. We shall take half of the coefficient of x, i.e., half
of b / a, square that, then add it to both sides of the equation. (It is added to both sides, of course, in order to
keep the equation in balance.) Taking half of a number is the same as doubling the denominator, right?
It’s not terribly obvious, but it is possible to factor the left side into perfect squares. At the same time, let’s
combine the fractions on the right side.
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Notice that I reversed the numerators on the right side. It can be verified that the factoring on the left side is
correct by simply multiplying things back.
Let’s stand back a moment and see what we have. The left side is a variable term squared. The right side
is a constant. Why? The variable in this problem is x. Letters a, b and c represent constant numbers, so since
everything on the right side only includes constants, the right side calculates to be a constant. Thus we can finish
solving by using the square root method.
Note that we were able to simplify the denominator to 2a.
It is a simple matter to finish isolating x by subtracting the b / 2a term from both sides. Since there is a
common denominator, the terms on the right side are traditionally written over the common denominator. Thus
we have derived the quadratic formula.
I do not require that my students learn the completing the square method for solving quadratic equations,
although it is, of course, perfectly acceptable. I learned this stuff years ago in high school and we spent about a
week drilling completing the square each day. We don’t have that time luxury in college, and since the quadratic
equation is faster to learn, it is more practical to focus on it. Thus we will leave completing the square behind in
this book with the observation that the quadratic formula is not some magical formula, but is derived by using
completing the square.
The quadratic formula is, in my opinion, one of the very few formulas in mathematics worth memorizing,
especially if you are headed to calculus. Professors will simply expect you to know it. Memorizing it is a simple
matter of writing it down each time you use it. Soon you will know it without any other effort.
_________________________________________________________________________________________
Topic 5: Applying the quadratic formula.
Solve:
27
The first order of business is to arrange this equation in descending order. We should add 4x to each side
and organize the terms in descending powers, i.e., x2 term first, followed by the x term, and then the constant.
Descending powers order ensures that we don’t accidentally mix up the proper numbers for a, b and c in the
quadratic formula.
Now we can safely isolate the constants. I think it is a good idea to write these down.
I also think it a good idea to write down the quadratic formula before you plug numbers in. The more you have
in front of your face instead of stored in your head, the less chance there is for careless errors.
rors.
Notice the generous use of parentheses when I plug in the constants. These can help you avoid sign er-
When you can simplify the square root, you should. I expect you to simplify where possible and I will count off
if you do not. We may also be able to simplify the entire fraction.
Recall that simplifying algebraic fractions involves factoring and canceling common factors.
Here is our final answer.
Here’s a good question: Is it possible to have a quadratic equation in which, while solving, the number
under the square root is negative? Indeed it is. One such example is
. The solution to this
problem is
. You may have run across such numbers—complex numbers—before. These are
introduced in the next section. At this point in your textbook these problems are classified as “no real solution.”
28
5.
Solve:
6.
Solve:
_________________________________________________________________________________________
Topic 6: Calculating approximate solutions.
I require my students to also be able to calculate the approximate solutions to quadratic equations on their
calculators. An answer such as
is perfectly fine for algebraic purposes. It is an exact, symbolic representation of the two solutions to a particular
quadratic equation. But has little practical value if what you need are decimals.
There are several important things to know about finding approximate decimal representations to these
quadratic solutions.
How does your calculator handle square roots?
Older style scientific calculators require you to punch in the number and then hit the square root button.
For example, to find
, you would first enter 17, then punch the square root button. Graphing calculators,
plus more recent scientific calculators that show the operations on the screen, have you enter the square root first
followed by the number. You should be aware of which order your calculator requires.
Be aware of the order of operations; use parentheses.
Your calculator is a miniature, programmed computer, so it operates under specific, rigid rules. Consider
the following simple calculation:
. If you type in 2 + 6 ÷ 4 = you will get the wrong answer, 3.5. The
order of operations requires you to complete the addition on top before dividing since the top numbers are grouped
together as if they are enclosed by parentheses. So you need to think of the calculation as
. Since you
should have a scientific calculator for this course, it will have both a left and right parenthesis button. So you
would enter this calculation as ( 2 + 6 ) ÷ 4 =, which will give the correct answer of 2.
Real-numbered and complex-numbered solutions are organized and calculated differently.
This is a source of real trouble with many students, who seem to forget the rules for combining like terms.
We will tackle complex numbers in the next section, numbers which look something like 2 + 3i. Would you
combine 2 + 3i to get 5i ? I hope not. Very few students would, but quite a few would wrongly combine
0.375 + 1.053i and get 1.428i. There’s something about those decimals that causes many students to forget
the basics of combining like terms.
Thus, pay close attention to the differences in how we calculate complex solutions and real-numbered
solutions.
Approximate
29
, rounding to three decimal places.
There is nothing magical about three decimal places, but that seems to be sufficient precision for most of
the quadratic equations I expect you to solve.
It won’t do you a bit of good to read this unless you are also trying this on your calculator.
This being a real-numbered solution, I will first separate the two answers.
We separate them because we need to calculate each separately. When calculating, it is a good idea to think of
the numerators as being enclosed by parentheses.
The first one calculates to 0.561552812809. Since we are rounding and the first decimal after the third
decimal place is 5 (or more), we round up. The second answer also rounds up for the same reason.
I like to use the approximately equal sign, ≈ , so that my readers know that I have approximated.
________________________________________
Approximate
, rounding to three decimal places.
Like the last one, we should separate the answers and use parentheses. You can do this mentally, of course,
but until you’ve had a lot of experience, you’ll be more prone to errors that way.
This time the second answer rounds up, but the first does not.
7.
Solve:
. Give both the exact symbolic answers and the approximate decimal
answers, rounded to three decimal places.
8.
Solve:
. Give both the exact symbolic answers and the approximate decimal
answers, rounded to three decimal places.
_________________________________________________________________________________________
30
Topic 7: A basic quadratic application.
The length of a rectangle is 3 inches longer than the width. The diagonal of the rectangle is 5 inches longer than
the width. What are the dimensions of this rectangle? (Round to two decimal places.)
Let’s start by defining the common value to these statements, the width.
Here’s the diagram.
The key to this problem is noticing that a right triangle is involved. This almost always implies that the Pythagorean Theorem, which relates the sides of a right triangle, is needed. Most students know the Pythagorean
Theorem by the time they get this course, but in case you need a refresher, here it is.
Pythagorean Theorem: A triangle is a right triangle if and only if it has three sides of
length a, b and c such that
. In this case, c is the length of the hypotenuse, the side opposite the right angle.
By applying the Pythagorean Theorem to our problem we get an equation to solve.
It is a good idea to write out squares of binomials first before expanding them. Otherwise you may overlook
the middle terms from “foiling.”
This equation cannot be solved by factoring, so we use the quadratic formula.
31
Since we are more interested in the decimal approximation there is no need this time to simplify the square
roots.
tion.
One of these answers is obviously false: the negative answer. This is an example of an extraneous solu-
Always communicate the answer. Don’t forget the measurement units.
9.
The parallel sides of a trapezoid are 4 feet and 8 feet longer than the base. The area of the trapezoid is
25 square feet. How wide is the base? (Round to two decimal places.)
_________________________________________________________________________________________
Topic 8: Another quadratic application.
A cylindrical can is to be 8 inches tall and constructed out of 240 square inches of metal. Find the radius and
the volume, accurate to two decimal places.
What formulas will we need to work this problem? Obviously we need volume because that is what we
are seeking.
We also need surface area because one piece of information is expressed in terms of area (square inches).
Of course, we’ll be using r for radius and h for height.
We know nothing yet from which to calculate volume, but we do know something about the area, so let’s
start there. Since the height is 8 inches, we can replace h with 8 and state the area of the can, both as 240 square
inches, and in terms of the area formula.
Because we can, it will be a little easier to divide both sides by 2 in order to shrink the numbers. We should
also bring the 120 to the other side of the equation since this equation is a quadratic in terms of r.
32
To solve this quadratic equation it will be easier to leave π (“pi”) in symbolic form. When it comes time
to use the calculator you can then use the π button on your calculator, which has π stored accurately to as many
decimal places as your calculator holds. If you don’t have a π button then you must not have a scientific calculator.
Can you tell which answer is extraneous? It’s not possible to have two different cans like this, is it? Let’s
write out the results.
The one on the right is extraneous, isn’t it? That one calculates as a negative radius, which is impossible,
so we’ll throw it out and calculate the one on the left.
This will be a tricky one to enter accurately on a calculator. Here are where the parentheses belong.
My calculator shows the following result.
We will round this off to two decimal places when communicating the answer, but for now it is a good
idea to keep all the decimal places in your calculator to use in calculating the volume. Avoid rounding numbers
in the midst of calculations, else you can often create significant errors when the error created by rounding is
magnified by certain calculations.
With this decimal number on the calculator screen you’ll want to square it, and then multiply by 8π in order
to complete the volume calculation. Here is what I get on my calculator.
33
Now we can round off and present the answers.
_________________________________________________________________________________________
1.
2.
3.
34
4.
5.
6.
7.
8.
35
36
9.

Chapter 1 — Equations and Inequalities

Transcription

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