Equations and Inequalities: Plain Text for Non-Mathematicians (essentials) [1st ed. 2021] 3658327197, 9783658327194

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Guido Walz

Equations and Inequalities Plain Text for Non-Mathematicians

essentials

Springer essentials

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More information about this subseries at http://www.springer.com/series/16761

Guido Walz

Equations and Inequalities Plain Text for Non-Mathematicians

Guido Walz Darmstadt, Germany

ISSN 2197-6708 ISSN 2197-6716 (electronic) essentials ISSN 2731-3107 ISSN 2731-3115 (electronic) Springer essentials ISBN 978-3-658-32719-4 ISBN 978-3-658-32720-0 (eBook) https://doi.org/10.1007/978-3-658-32720-0 The translation was done with the help of artificial intelligence (machine translation by the service DeepL.com). A subsequent human revision was done primarily in terms of content. © Springer Fachmedien Wiesbaden GmbH, part of Springer Nature 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Responsible Editor: Iris Ruhmann This Springer imprint is published by the registered company Springer Fachmedien Wiesbaden GmbH part of Springer Nature. The registered company address is: Abraham-Lincoln-Str. 46, 65189 Wiesbaden, Germany

What You Can Find in This essential

• Formulas for solving linear and quadratic equations • Methods for the solution of root and fractional equations • Strategies for the solution of linear inequalities and fractional inequalities

v

Introduction

Solving equations is an omnipresent problem when dealing with problems from natural sciences, technology or even economics. It does not necessarily have to be about higher mathematics, even simple equations like linear or quadratic ones can cause problems. These types of equations will be dealt with in the following. It is not just a matter of merely solving them, but also of dealing with equations that at first seem inaccessible, but which, after clever remodelling, turn out to be easily controllable and, above all, solvable. This includes the fractional and root equations, which are treated in the following pages. Not quite as often as equations, but still often enough one encounters the problem of having to solve an inequality. These are also discussed below, especially linear inequalities and fractional inequalities leading to linear ones. This booklet will help you by presenting ready-made rules for solving the above equations and inequalities and by giving you the opportunity to practice them using numerous selected examples. Since the subtitled text is explicitly (also) aimed at non-mathematicians, it is deliberately written in generally understandable language so as not to discourage readers by using excessive technical language; after all, the subtitled text is also supposed to be “plain text.” For example, we will speak of “superscripts” instead of “exponents,” and likewise of “unknowns” instead of “variables.”

vii

Contents

1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

ix

1

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Summary

1.1

What Does an Equation Look Like Anyway?

An equation consists of (at least) two expressions connected by an equals sign, thus it has the form Expression1 = Expression2 For example 2+2=4

(1.1)

an equation which is not really anything to criticize, except of course that it is boring. You can also combine several expressions into a single equation—hence the “at least” in brackets in the first line—for example, 2 + 2 = 3 + 1 = 4. This is helpful when several forming steps are carried out in succession, for example (3 + 2) · (9 − 3) = 5 · 6 = 30.

© Springer Fachmedien Wiesbaden GmbH, part of Springer Nature 2021 G. Walz, Equations and Inequalities, Springer essentials, https://doi.org/10.1007/978-3-658-32720-0_1

1

2

1

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But even that is still relatively boring. It becomes more interesting if the equation contains an unknown—in technical jargon also: Variable -, something like 2+x =4

(1.2)

sin(log(tan(x 42 ))  cos(x) . = 42 3 sinh(2x) + 17

(1.3)

or something creepy like

In these cases, the point is to find those numbers that can be used for x, so that the expressions on both sides of the equals sign are really the same. This is called solving the equation, and the x-values concerned are called the solutions of the equation. The summary of all solutions of an equation is often called the solution set of this equation. At the latest now it becomes clear why I described equations like (1.1) above as boring: There is nothing to solve here, the equation is simply right or wrong. I think we can agree that Eq. 1.2 has exactly one solution, namely x = 2. Before you despair now because you have no idea how to solve Eq. 1.3: There you are like me, it was just an extravagant example to keep the tension high. Chitchat The equal sign “=” was introduced in 1557 by Robert Recorde in his algebra textbook “The Whetstone of Witte.” It is supposed to symbolize two parallel lines, because in Recorde’s view there is nothing more equal than two parallel lines (“bicause noe 2 thynges can be moare equalle”) (Source: Lexikon der Mathematik).

1.2

What Can You Do with an Equation Without Changing Its Solution Set?

Well, first of all, any expression that occurs in an equation can be transformed at will, such as multiplying or summarizing, as long as this does not change its value. For example, you cannot tell at first glance whether the equation

1.2 What Can You Do with an Equation …

2 · 17 − 24 46 = (−5)2 − 5 2

3

(1.4)

is true—I certainly cannot. But if you calculate both expressions, you get on the left side 34 − 24 10 2 · 17 − 24 = = = 2, 5 5 5 on the right (−5)2 −

46 = 25 − 23 = 2. 2

Equation 1.4 is therefore “actually” only 2 = 2 and “is therefore true.” It can also be helpful in solving equations that contain an unknown: The equation x + 3x + (2 · 3)x + 7x = 17 looks quite strange; but when I summarize the left side, it just says 17x, so the equation is simplified: 17x = 17, and has the only solution x = 1. Most of the time, however, one will have to manipulate both sides of the equation to find its solution. For example, if you need to solve x + 3 = 7, it will be a good idea to subtract 3 on both sides of the equation. This will immediately give you the solution x = 4 to the initial equation, and that is certainly the solution to the initial equation. It won’t always be that easy, very often you have to change the equation several times to get to the goal. But what can you do with an equation without changing its solution set—returning to the title of the chapter again? Rule The following manipulations of an equation do not change its solution set and may therefore be used as often as desired: • Add or subtract the same number on both sides of the equation

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• Multiply both sides of the equation with the same number, which must not be zero • Divide both sides of the equation by the same number, which must not be zero You will find plenty of examples of this on the following pages, so I will limit myself to just one: The equation x 2 + 3x − 84 = x 2 + x.

(1.5)

I subtract on both sides x 2 + x and thereby obtain the already greatly simplified equation 2x − 84 = 0, and if you now add 84 on both sides and then divide both sides by 2, the solution is explicitly x = 42 there. And if you don’t believe me—which I could understand at any time—you can put this value into the initial Eq. (1.5) as a check; you get the same on both sides, namely 1806. Know-it-all info According to the above rule, there are no restrictions when adding or subtracting numbers, while multiplying and dividing with or by zero is not allowed. Why is this so? Well, adding or subtracting the number zero on either side of the equation does not change it in the slightest, so it is not surprising that this process is allowed. Whether it is very useful for solving the equation remains to be seen. Multiplying both sides with zero immediately transforms each equation into the new equation 0= 0. While this has the charm of being correct, it has the disadvantage that the solution of the original equation can no longer be determined from it, and this cannot be reversed. After all, dividing anything by zero is simply not allowed, it has nothing to do with solving equations in the strict sense.

1.3 Linear Equations

1.3

5

Linear Equations

Some types of equations occur repeatedly, so there are “ready-made” formulas for them; we will look at these in more detail on the following pages. I start with the linear equations. Example 1.1

A car driver drives for several hours at constant speed on a level road (in a mathematics book this is possible). Due to its gentle driving style, the car consumes only 5 L of petrol per 100 km distance, that is, 0.05 L/km. At the beginning of the journey the tank was filled with 40 L petrol. If you sum this up, you get the “petrol function.” b(x) = 40 − 0.05 · x, which indicates how many liters of petrol are still in the tank after x kilometers driven. After a few hours, the driver suspects that the tank could soon be empty, and he wants to calculate when this will be the case. How can he do that? Well, “empty tank” means nothing more than that the fuel function has a value of zero, so it is the equation 40 − 0.05 · x = 0

(1.6)

to solve. Because x is linear here, that is, without a superscript, root or other disturbing additions, this is called a linear equation. Definition 1.1 An equation of form mx + d = 0 with real numbers m and d, where m must not be zero, is called linear equation. You have already seen a first example of such a linear equation in (Eq. 1.6): There was m = −0.05 and d = 40. Further examples follow in a moment.

6

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Equations

Know-it-all info The requirement that m  = 0 is certainly sensible, because if it is m = 0, the equation is reduced to d = 0, and that is either wrong, if d is not zero, or meaningless. The nice thing about linear equations is that they always have one—and exactly one—solution, and that you can give it without too much calculation; and that is exactly what I want to do now. Rule The linear equation mx + d = 0 has the solution x1 = −

d . m

There is not too much to say about the derivation here: First deduct d on both sides of the equation and then divide by m. And there is the given solution formula. Know-it-all info With everything you do in your mathematical life (whatever that is exactly), you should ask yourself if you are not doing something strictly forbidden, such as dividing by zero or taking the square root of a negative number. Neither of these things happens here: The only thing that is divided here is the number m, which, according to the definition of the linear equation, must not be zero, and there is nothing to be seen of a root here for miles around.

Example 1.2

a) In the linear equation 2x + 5 = 0 we have m = 2 and d = 5. Consequently, their solution is 5 x1 = − . 2

1.3 Linear Equations

7

You might want to add that to the equation for verification purposes. You never know with me. b) The equation 3x − 7 = 0 has the solution x1 = −

−7 , 3

which, because of the good old rule “minus times minus equals plus,” can be written down even more nicely x1 =

7 . 3

c) The equation 2x + 5 = 47 − x you got to get your mind right a little bit first: Addition of x and subtraction of 47 on both sides turns it into the linear equation 3x − 42 = 0. It has the solution x1 = −

42 −42 = = 14, 3 3

and this value is therefore also the solution of the initial equation.  Example 1.3

We should finally come back to the driver from example 1.1 and calculate how far his fuel will go. For this we have to solve the equation 40 − 0.05 · x = 0

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Equations

solve. But with the new knowledge you have just acquired it is easy: Here m = −0.05 and d = 40, that is x1 = −

40 . −0.05

5 Now what? Well, you can, for example, rewrite the value −0.05 to − 100 and then apply tighter fractions: The good old rule: “You divide by a fraction by multiplying with its reciprocal” leads to

x1 = −

40 5 − 100

=−

4000 40 · 100 =− = 800. −5 −5

However, you can also use the calculator … In any case, you get the result that the gasoline is sufficient for a total of 800 km. I can’t think of much more about linear equations, so let’s go straight to the more sophisticated quadratic equations.

1.4

Quadratic Equations

1.4.1

First Examples

To say it right from the start: The term “quadratic equation” is not quite correct, strictly speaking, the equation itself is of course not quadratic (but rather broad), just as the electric locomotive driver is not powered by electricity. The name is rather due to the fact that in such equations the x is not only linear, but also has the superscript 2, that is, the equation contains the expression “x to the power of two” or “x squared.” But what exactly does a quadratic equation look like? First, an example from the world of text problems. Example 1.4

A rectangular plot of land has the area of 351 square meters and the perimeter 80 m. Is it possible to calculate the length of the sides of the plot with this information? To answer this question we should first give the two side lengths names, let’s call them x and y. Since the area of a rectangle is just the product of the

1.4 Quadratic Equations

9

side lengths, I know from the first specification that x · y = 351 is—whereby I have omitted the units in the best mathematical tradition. To determine the circumference of the rectangle, you have to take twice the length of the sides and add them up; thus 2x + 2y = 80. So now we have two equations with two unknowns; unpleasant thing. To reduce this to an equation with only one unknown, I first solve the second one after one of the unknowns, let’s say after y. This is done in two steps like this: First you divide both sides by 2, which results in x + y = 40, and then subtract x on both sides, resulting in y = 40 − x. In words: y is the same as 40 − x, and therefore I can replace y with 40 − x in the first equation without breaking anything. This results in x · (40 − x) = 351. If you now multiply out the left side and shift the 351 to the left side, you get the equation −x 2 + 40x − 351 = 0.

(1.7)

This is my first example of a quadratic equation. The answer to the question asked at the beginning, whether one can calculate the side lengths of the rectangle, is now quite clear: “It depends.” If you can solve quadratic equations, yes, otherwise no. A few pages further on at the latest, your answer will be “yes”. I have quite a few more examples for you, but first I should write down exactly what a quadratic equation is. It’s called a definition.

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Definition 1.2 • An equation of form ax 2 + bx + c = 0 with real numbers a, b, and c, where a must not be zero, is called a quadratic equation. • An equation of the form x 2 + px + q = 0 with real numbers p and q is called quadratic equation in normal form. A quadratic equation has a normal form if the coefficient, that is the number before x 2 equals 1 (and is therefore not explicitly written down). Why then renaming the other two numbers p or q, you shouldn’t ask yourself or me, that’s what has become common practice. Example 1.5

a) The equation 2x 2 + 3x − 2 = 0

(1.8)

is a quadratic equation with a = 2, b = 3 and c = −2. b) The equation −x 2 + 40x − 351 = 0 from the input example is a quadratic equation with a = −1, b = 40 and c = −351. It has no normal form, which is x 2 due to the inconspicuous minus sign. c) The equation x 2 − 2x − 3 = 0 is a quadratic equation in normal form with p = −2 and q = −3. d) The equation

(1.9)

1.4 Quadratic Equations

11

3x 2 −

√

x +5=0

is not a quadratic equation (because of the root).  Know-it-all info The requirement that a  = 0 is also useful here, because if it is a = 0, the equation is reduced to bx + c = 0, so a linear equation, and we have already looked at linear equations further ahead. The other two coefficients b and c (or p and q), on the other hand, may well be zero; the following examples show that this may even have advantages. Example 1.6

a) The poorest quadratic equation I can think of is x 2 = 0.

(1.10)

So here b and c are zero at the same time. At least it can be easily released: Here we look for a number x, which squares, that is, multiplied by itself, gives zero. However, this can only be achieved by a number, namely x = 0. So this is the only solution of the quadratic Eq. 1.10. b) Not much more demanding is the equation 2x 2 − 8 = 0.

(1.11)

So here a = 2, b = 0 and c = −8. The essential point is that b = 0, because it makes it easy to solve the equation: Dividing by 2 and then adding the constants 4 on both sides forms (1.11) to convert into x 2 = 4.

(1.12)

Surely you have already seen that one solution of this equation is x 1 = 2, but beware! There is also a second solution, namely that x2 = −2 because of the rule already quoted, minus times minus equals plus (−2) · (−2) =

12

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Equations

4. Equation 1.12 and therefore also Eq. 1.11 has two different solutions, namely x1 = 2 and x2 = −2. c) Also for the Equation x2 + 9 = 0 b = 0, but this does not help the solution: If you put 9 on the right side, you get x 2 = −9. A solution of this equation (and thus also of the initial equation) would therefore be a number x, which multiplied by itself results in something negative. But there is no such thing, at least not in the range of real numbers in which we are moving here. Therefore, this equation, as harmless as it looks, is unsolvable. d) You should also see c = 0 an example of the case; how about this: x 2 + 3x = 0.

(1.13)

Now comes a little trick that you can always use if you want c = 0: I put the factor x on the left side outside the brackets and get x · (x + 3) = 0. Now, a product is zero exactly when at least one of the factors is zero. Therefore the expression x · (x + 3) is zero if either x = 0 or x = −3. And thus Eq. 1.13 has the two solutions x1 = 0 and x2 = −3. Every quadratic equation can be brought into normal form by dividing both sides of the equation by a (whereby nothing changes on the right side, because zero by a is zero, so a can be as large as it wants). The concrete calculation rule for this is as follows: Rule Any quadratic equation ax 2 + bx + c = 0 can be brought into normal form

1.4 Quadratic Equations

13

x 2 + px + q = 0 by setting: p=

b c and q = . a a

As I said, nothing else happened here but that every single summand was divided by a. For example, the normal form of Eq. 1.8 is: x2 +

3 x − 1 = 0. 2

(1.14)

Just as further ahead in the solution of the linear equation, it must be emphasized that one may divide by a without restraint, since it is assumed to be unequal to zero. Since every quadratic equation can now be brought into normal form, no formula would be needed to determine the solution of a general quadratic equation; it would be sufficient to be able to solve the quadratic equation in normal form. This is quite true, and quite a few mathematicians—myself included—are basically only concerned with equations in normal form and their solution by the so-called p-q-formula, which I will show you below. For the sake of completeness, and because the formula for the solution of the general equation has such a nice name, I want to introduce it briefly in the next section.

1.4.2

The Midnight Formula

The following infobox gives complete information about how many solutions a quadratic equation can have and how to calculate them if necessary. In particular, it contains the so-called midnight formula (1.15); it is also known under the very profane and far from beautiful name a-b-c formula. Know-it-all info A quadratic equation ax 2 + bx + c = 0 has either no solution at all, or one or two different solutions.

14

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To find out which of these situations exists and to calculate the solutions if necessary, one first determines the number D = b2 − 4ac. • If D is negative, the quadratic equation has no solution. • If D is zero, the quadratic equation has exactly one solution, namely x1 = −

b . 2a

• If D is positive, the quadratic equation has two different solutions x 1 and x2 , which can be calculated using the midnight formula √ √ −b + D −b − D and x2 = . x1 = 2a 2a Usually this is summarized in the short form x1/2

√ −b ± D = 2a

(1.15)

together. Chitchat The midnight formula is so called because many people think that every student must be able to recite it by heart when you wake him up at midnight. As I said above, I don’t necessarily agree with this opinion, because the easier to remember p-q-formula does the job. I think only the math-enthusiasts among you can really enjoy formulas like (1.15), and you probably spontaneously think of some things you’d rather do at midnight than recite that formula. So I’ll just give you an example and then go straight to the p-q-formula, which is at least a little bit more compact than the midnight formula. Example 1.7

a) Let’s start by looking at the equation

1.4 Quadratic Equations

15

2x 2 + 3x − 2 = 0

(1.16)

from example 1.5. Here a = 2, b = 3 and c = −2, which returns the value D = 32 − 4 · 2 · (−2) = 9 + 16 = 25, and since 25 is positive, we know that the equation has two solutions. The midnight formula provides x1 =

√ √ −3 + 25 −3 − 25 2 1 −8 = = and x2 = = = −2. 2·2 4 2 2·2 4

Equation 1.16 therefore has the two solutions x1 = 21 and x2 = −2. If you do not believe this—which I would understand immediately, because I am always making miscalculations—you can easily check it by inserting these values into the left side of the equation; try it, I will wait here as long as it takes. b) For the equation which is only slightly different from (1.16) 2x 2 + 3x + 2 = 0

(1.17)

results in the value D = 32 − 4 · 2 · 2 = 9 − 16 = −7, which is a negative number. The actually quite harmless looking Eq. 1.17 is therefore unsolvable.  Example 1.8

I had promised at the end of example 1.4 that you would be able to calculate the side lengths “a few pages further” by solving Eq. 1.7; and that is now ready. The equation to be solved is −x 2 + 40x − 351 = 0.

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So here a = −1, b = 40, and c = −351, and so D = 402 − 4 · (−1) · (−351) = 1600 − 1404 = 196, is positive. Therefore the equation has two different solutions, namely x1 =

√ −40 + 196 −40 + 14 = = 13 −2 −2

and √ −40 − 196 = 27. x2 = −2 The respective other side length y is obtained from the correlation y = 40 − x determined in the example, it is y1 = 40 − x1 = 27 and y2 = 40 − x2 = 13. And now you can also see that the side lengths of the rectangle are in fact clearly determined, they are 13 and 27, because which side I call x and which I call y is of course my own artistic freedom. That’s it with the general quadratic equations, I turn now to the p-q-formula (and you’re bound to do so, after all you paid money for this book).

1.4.3

The p-q-Formula

Without too much preface, I would like to present the p-q-formula for solving quadratic equations in normal form. Since you have already seen that any quadratic equation can easily be brought into normal form, this means nothing else than that you can use this formula to solve any quadratic equation. Rule A quadratic equation in normal form x 2 + px + q = 0

1.4 Quadratic Equations

17

has either no solution at all, or one or two different solutions. To find out which of these situations exists and to calculate the solutions if necessary, one first determines the number d=

p2 − q. 4

• If d is negative, the quadratic equation has no solution. • If d is zero, the quadratic equation has exactly one solution, namely p x1 = − . 2 • If d is positive, the quadratic equation has two different solutions x 1 and x2 , which can be calculated using the p-q-formula

x1 = −

p √ p √ + d and x2 = − − d. 2 2

Usually this is summarized in the short form x1/2 = −

p √ ± d 2

(1.18)

together. Yes, I know that the whole thing is very similar to the midnight formula, and I also admit that I have been working with copy and paste here, one has to move forward. But please note: I would like to recommend the p-q-formula as a rule, because in my opinion and experience you should use it; the midnight formula, on the other hand, I have formulated as know-it-all info. Chitchat In painting, music, architecture, and many other fields, the golden ratio plays an important role. In the language of geometry it can be described as follows: A rectangle is constructed according to the golden ratio, when the ratio of the longer side to the shorter is the same as the sum of the two side lengths to the longer. Any more questions? Probably.

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Fig. 1.1 A rectangle

Surely a look at Fig. 1.1. Here I have drawn a rectangle, the longer side named x and the shorter side y. The above requirement now means in formulas x+y x = . y x

(1.19)

Since it only depends on the length ratio of the two sides, I may normalize one of them and do so by setting the length of the shorter side y = 1 to 1. That makes (1.19) the already much more friendly looking equation x x +1 = . 1 x If we multiply by x and leave out the already unnecessary denominator 1, we get the equivalent equation x 2 = x + 1, and if you now bring everything to the left side, you get the quadratic equation in normal form x 2 − x − 1 = 0.

(1.20)

1.4 Quadratic Equations

19

The value d from the p-q-formula is equal to d=

1 1 5 − (−1) = + 1 = , 4 4 4

is positive. Thus there are two different solutions of the quadratic Eq. 1.20, namely 1 x1 = + 2



5 1 ≈ 1.618 and x2 = − 4 2



5 ≈ −0.618. 4

Since negative side lengths are rather unusual for rectangles, the only reasonable solution to the problem is x1 = 1.618.... In case you have forgotten which problem we have actually solved here (which I could well understand), here again in summary: The question was what the ratio of the two sides of a rectangle constructed according to the golden ratio must be; and the answer is: The longer side must be about 1.618 times as long as the shorter one. By the way, the rectangle in Fig. 1.1 is actually drawn according to this principle. Such figures are considered to be particularly beautiful. (No, I don’t want to argue about that now.) For example, the Parthenon on the Acropolis in Athens is built according to the golden ratio—although it is not known whether the builders applied this construction principle on purpose or whether they chose a “beautiful” ground plan rather intuitively. But now enough of chitchat, here are a few examples of the p-q-formula, because to learn mathematical contents, there is nothing better than practicing, practicing, and in case of doubt practicing again. Example 1.9

a) The quadratic equation to be solved is x 2 + x − 2. So here p = 1 and q = −2. So there is d=

1 1 9 − (−2) = + 2 = , 4 4 4

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and since 49 is positive, two different solutions of the quadratic equation can be expected. These are 1 x1 = − + 2



9 1 3 1 3 = − + = 1 and x2 = − − = −2. 4 2 2 2 2

b) The second equation we want to look at is x 2 − 3x = 0.

(1.21)

You may remember example 1.6 d and therefore see that you can exclude x on the left side and thus convert the equation into the form x(x − 3) = 0. From this, in turn, it can be directly seen that the equation has the two solutions x2 = 0 and x1 = 3, because for this one of the factors on the left side becomes zero. But that’s exactly what I don’t want to know for the time being (older mathematicians can be quite stubborn), but without further thought I will throw the p-q-formula on (1.21). This is not quite as pointless as it might seem: In a stressful situation, such as a test, you may not see the decomposition possibility just shown, and then it helps if you can hold on to something as secure as the p-q-formula. So here p = −3 and q = 0, and so d=

(−3)2 9 − 0 = > 0. 4 4

The two solutions of the equation to be expected are −3 x1 = − + 2



9 −3 3 3 = + = 3 and x2 = − − 4 2 2 2



9 3 3 = − = 0, 4 2 2

in full accordance with the result obtained at the beginning. c) Now we look at the quadratic equation x 2 − 6x + 9 = 0

(1.22)

1.4 Quadratic Equations

21

I think I can save myself the trouble of explicitly naming p and q again, you’ve got that up to now. For d I get the value d=

(−6)2 36 −9= − 9 = 0. 4 4

Equation 1.21 thus has only one solution according to the p-q-formula, namely x1 = −

−6 = 3. 2

Know-it-all info Perhaps you know the binomial formula, even though I have not mentioned it in this text with any word so far. If you do, you can apply it once to the left side of Eq. 1.22, and you will get the expression (x − 3)2 , the equation is equivalent to (x − 3)2 = 0. In this form, however, you can see with the naked eye that the equation has x = 3 as the only solution. So we have found two different ways to solve (1.22); there are worse things to do. d) No one has claimed that the numbers appearing in an equation must always be integers. Here are the following two examples. To solve the quadratic equation x 2 − 3x +

11 = 0. 4

(1.23)

Here d=

1 9 11 − =− , 4 4 2

is negative. That is why the “to solve is” such a thing: The square Eq. 1.23 is not solvable. e) One should not stop like this, so at the end of this example, therefore, a solvable equation. But without integer coefficients:

22

1

x2 +

5 1 x − = 0. 12 6

Equations

(1.24)

So here we will have to deal a little with fractions. Don’t worry, I’m with you—if it’s not this very fact that worries you. The worst thing here is probably to calculate the value d, so let’s get it over with. First… d=

1 25 − (− ), 144 · 4 6

because 52 = 25 and 122 = 144. The two minus signs at the back “lift themselves away,” as my children would say, and it might also be a good idea to bring the whole thing down to a common denominator 144·4 = 576. This will result in d=

25 1 25 96 121 + = + = 576 6 576 576 576

Whatever this may be exactly, it is in any case a positive number, and therefore Eq. 1.24 has two solutions. These are 5 x1 = − + 24



5 11 6 1 121 =− + = = 576 24 24 24 4

and 5 x2 = − − 24



121 5 11 16 2 =− − =− =− . 576 24 24 24 3

By the way, in case you didn’t immediately see that the root of 576 is 24, you are just like me. But that doesn’t matter, just put the calculator on it! In exercises or exams, you usually get smooth values, because your lecturers don’t want to constantly recalculate “crooked” values. 

1.4 Quadratic Equations

1.4.4

23

Quadratic Equations Which Are Not Recognizable at First Glance

Sometimes quadratic equations appear hidden, that is, an equation that may initially appear much more complicated can be reduced to a quadratic equation. In this short section, I will introduce three situations in which such things typically occur. I will show the first situation with an example: The equation to be solved is x(x + 1)(x + 2) = x 3 + 2x 2 − 1. At first sight hardly feasible, because here even the superscript 3 appears, so the equation seems to be far from being a quadratic equation. But it just seems so; if you multiply the left side and sort by x-power, you get the equation x 3 + 3x 2 + 2x = x 3 + 2x 2 − 1. (Don’t despair: I didn’t do this in one step either, a little extra calculation was necessary here). Now you subtract x 3 on both sides, which results in 3x 2 + 2x = 2x 2 − 1. This looks much more friendly, because the superscript 3 has completely disappeared. Now I bring everything to the left side of the equation by subtracting 2x 2 on both sides and adding 1; I get x 2 + 2x + 1 = 0, thus a quadratic equation of purest form, and even thankfully already in normal form. It has the only solution x1 = −1, as you can see for example with the help of the p-q-formula. And why does this equation not look as a quadratic one, as it was said in the title? Because here, first of all, even greater superscripts than the 2 appear and the equation therefore does not look like a quadratic one at first. But since the expressions with these larger superscripts are fortunately balanced on both sides, they are omitted when the equation is smoothed and only a quadratic equation remains. Here is another example. Also the equation

24

1

Equations

1 3 2 x(x 2 − 1)(3 − 2x) = x 3 (1 − x) 2 2 3

(1.25)

looks very inaccessible at first glance. But this is deceptive, as you can see by multiplying and then sorting by x-power: The left side can be transformed in two careful steps into 1 1 3 1 3 x(x 2 − 1)(3 − 2x) = ( x 3 − x)(3 − 2x) = x 3 − x − x 4 + x 2 , 2 2 2 2 2 the right one to 3 3 x − x 4. 2 The Eq. 1.25 formulated at the beginning is therefore equivalent to 3 3 3 3 x − x − x 4 + x 2 = x 3 − x 4. 2 2 2 If you look sharply, you will see that the complete right side also appears on the left side and can therefore be removed on both sides. It remains 3 − x + x 2 = 0. 2 But this is a quadratic equation (the purists can still bring that x 2 forward to make complete agreement with the definition), it has the two solutions x1 =

3 and x2 = 0, 2

and if we have not made any conversion errors along the way, this also applies to the initial Eq. (1.25). Try it out by inserting these two values into the output equation. A second situation, in which quadratic equations can appear hidden, we have already seen above in the conversation about the golden ratio. There the equation x x +1 = 1 x a so-called fractional equation, as a variant of the quadratic equation

1.4 Quadratic Equations

25

x2 − x − 1 = 0 has been exposed. Another, similar example of this kind is the equation x − 41 =

40 . x −2

(1.26)

To smooth this out, first multiply by the denominator (x − 2), which leads me to the equation (x − 41)(x − 2) = 40, so we get x 2 − 43x + 82 = 40 If you subtract 40 on both sides, you get the quadratic equation in normal form x 2 − 43x + 42 = 0. Don’t be afraid of big numbers! The number d usually defined for the p-q-formula here d=

1849 1849 168 1681 432 − 42 = − 42 = − = , 4 4 4 4 4

is positive. Thus the quadratic equation has two solutions, and without much ado I calculate these as  43 41 43 43 41 1681 + = + = 42 und x2 = − = 1. x1 = 2 4 2 2 2 2 And these two values are in fact also solutions of the initial Eq. (1.26), as one checks by inserting the respective value on both sides of the equation—I will refrain from the “effortless” that is otherwise usual at this point. Chitchat By the way, equations such as the one just discussed can also show you the strength of the mathematical “formula language,” which is frowned upon by most people

26

1

Equations

and makes some people downright afraid of mathematics. Completely unnecessary, because formulas only serve to make the statements, here the task, more precise and thus clearer. For if you wanted to put Eq. 1.26 into words, you would have to say something like “I am looking for numbers which have the property that they give the same result if you subtract 41 from them on the one hand and subtract 2 from them on the other hand and then divide 40 by the value just received.” You don’t even understand the task at hand, let alone the solution. And a third and last example will show you that behind fractional equations there are often “only” quadratic equations. Example 1.10

To be solved 1 1 1 = − . x x +1 x −1

(1.27)

In order to get rid of all three denominators here, I unfortunately have to multiply by your product, since obviously none of the three denominators is contained as a factor in any of the others. This gives me an intermediate result x(x + 1)(x − 1) x(x + 1)(x − 1) x(x + 1)(x − 1) = − , x x +1 x −1 and after shortening the factors that occur simultaneously in numerator and denominator, (x + 1)(x − 1) = x(x − 1) − x(x + 1). Now it’s time to start multiplying out again: I get x 2 − 1 = x 2 − x − (x 2 + x), so x 2 − 1 = −2x or

1.4 Quadratic Equations

27

x 2 + 2x − 1 = 0. This quadratic equation also has two solutions, namely x1 = −1 +



1 − (−1) = −1 +

√ √ 2 and x2 = −1 − 2.

 A quadratic equation can also be hidden under a root, so to speak, which brings us to the topic of root equations at the very end of this chapter. A first example is the equation 

x 2 + 2x + 6 + 3 = 0.

(1.28)

To get closer to the solution of this equation, you have to somehow get rid of the root, and since the natural enemy of root extraction is squaring (that is, multiplying by itself), this is the one to use. However, if you apply this operation to the equation in the form given here, you have to apply the binomial formula on the left (or multiply it out on foot) and you get   ( x 2 + 2x + 6 + 3)2 = (x 2 + 2x + 6) + 6 · x 2 + 2x + 6 + 9. You can now summarize this further as you like, you will not get rid of the root. It is much better to first convert the initial Eq. (1.28) into the form 

x 2 + 2x + 6 = −3

to bring, and then to square. This immediately delivers x 2 + 2x + 6 = 9, so the quadratic equation x 2 + 2x − 3 = 0.

(1.29)

I hope you haven’t completely forgotten the p-q-formula about all the transformations of fractions and roots. You can now apply it to (1.29) and get the two solutions

28

1

x1 = −1 +

Equations

 1 − (−3) = −1 + 2 = 1 and x2 = −1 − 2 = −3.

But if you sit back contentedly and are happy to have found two solutions to the equation, you will be bitterly disappointed. With root equations it can happen that one finds solutions of the transformed equation—in this case the quadratic equation—which, however, do not fulfill the initial equation to be solved. These solutions are then called mock solutions. And that’s what’s happening here: If you insert the two values x 1 = 1 and x2 = −3 in (Eq. 1.28), you get both times the expression √ 9 + 3 = 0, who, by any stretch of the imagination, is not right. So we are dealing with two mock solutions, and the initial equation itself has no solutions at all. Know-it-all info Why is it possible that, when solving root equations, mock solutions can occur, that is, solutions of the transformed equation, but which are not solutions of the initial equation? Well, this is because interim squaring can change the solution set of the equation, because squaring is not only the natural enemy of root extraction, but also the absolute killer of negative signs; nobly formulated: Squaring is not an equivalence transformation. A very simple example already shows this: The equation −2 = 2 is certainly not correct even with the most generous interpretation of algebra. But if I square both sides of the equation, I get the correct equation 4 = 4. In order not to leave you quite so demotivated from this chapter, I will now show you two examples of root equations that can really be solved. Example 1.11

a) The first example is x−

√

210 + 37x = 0.

(1.30)

1.4 Quadratic Equations

29

One should always try to isolate the root on one side of the equals sign and bring everything else to the other side. It’s not always possible, but here it’s very easy and results in x=

√

210 + 37x.

Squaring both sides provides x 2 = 210 + 37x, thus the quadratic equation in normal form x 2 − 37x − 210 = 0. I have neglected the term d in the last examples. Here I calculate it again explicitly and find d=

372 1369 1369 840 2209 − (−210) = + 210 = + = . 4 4 4 4 4

Now this is the place where you think you have made a mistake or the person posing the problem has used nasty crooked numbers, but it only seems so: 2209 is a square number, to be more precise 2209 = 472 , and therefore the quadratic equation has the nice “smooth” solutions 37 x1 = + 2



2209 37 47 37 47 = + = 42 and x2 = − = −5. 4 2 2 2 2

To make sure that we didn’t work for nothing again and calculated only dummy solutions, I put these values in the left side of (1.30) and get 42 −

√

210 + 37 · 42 = 42 −

√ 1764 = 42 − 42 = 0;

so x1 is actually a solution of the initial equation. It does not look so good with x2 = −5, inserting this in (1.30) gives here −5 −



210 + 37 · (−5) = −5 −

√

25 = −5 − 5 = −10,

so not zero. x2 = −5 is therefore a mock solution.

30

1

Equations

b) One should solve 

2x 2 + x + 1 − x = 1.

(1.31)

Again, it is strongly recommended to isolate the root on the left side by adding x on both sides; this will result in 

2x 2 + x + 1 = x + 1,

an equation that can easily be squared and thus the following can be obtained: 2x 2 + x + 1 = (x + 1)2 , so 2x 2 + x + 1 = x 2 + 2x + 1. If you now bring everything to the left side, you get the clear quadratic equation x 2 − x = 0. Either with the p-q-formula or with the help of the decomposition one finds the two solutions of this equation x 2 − x = x(x − 1): x1 = 1 and x2 = 0. I hardly dare, but let us nevertheless insert these values into the initial Eq. (1.31). For on the left side we get x1 = 1 √

2+1+1−1=

√ 4 − 1 = 2 − 1 = 1.

x1 is therefore actually a solution. Encouraged in this way, I now take x2 = 0 and use it. I get √ √ 0 + 0 + 1 − 0 = 1 = 1.

1.4 Quadratic Equations

31

So solution x2 is also solution, so now we have also seen a root equation which has no mock solutions.  And that’s a happy ending to this chapter, don’t you think?

2

Inequalities

2.1

What Kind of Inequalities Are Meant Here?

The last chapter was about equations, that is, statements in which two or more expressions are connected by an equals sign and thus represent the same thing. One might now think that an inequality consists of two or more expressions connected by the inequality sign “ =.” In the literal sense of the word, this would also be correct, but things like 3  = 4 are not really sparkling, and also “inequalities” with unknowns, something like x 2 + 2x − 15  = 0

(2.1)

don’t bring much new. Here we are looking for all numbers x, which do not fulfill the corresponding equation x 2 + 2x − 15 = 0. But for this you can simply solve this quadratic equation using the p-q-formula (Attention: learning objectives!) and find x1 = 3 and x2 = −5. Thus, the “inequality” (2.1) is solved by all numbers which are not equal to 3 or −5. In the case of an inequality in the mathematical sense, however, one wants to know it a little more precisely and asks which of the two expressions is smaller than the other or for which values of the unknown x this is the case. This is ultimately due to the fact that the numbers we know in this way can all be arranged on a straight line, the so-called number line, and this means that of two different numbers (or more generally: expressions) one is always smaller than the other and can therefore be found further to the left than the other on the usual number line (Fig. 2.1).

© Springer Fachmedien Wiesbaden GmbH, part of Springer Nature 2021 G. Walz, Equations and Inequalities, Springer essentials, https://doi.org/10.1007/978-3-658-32720-0_2

33

34

2

Inequalities

Fig. 2.1 The number line

You want to be a little more specific? You’re welcome:  Definition 2.1 An inequality consists of (at least) two expressions connected by one of the characters < (“smaller”), ≤ (“smaller or equal”), > (“larger”), or ≥ (“larger or equal”). Example 2.1

For example 42 ≥ 17 is such an inequality, although not very exciting, but x + 2 < 13

(2.2)

is also an inequality, namely one that contains an unknown and therefore must be solved. What we are looking for here are those numbers x with the property: If you add 2 to x, the result is still smaller than 13. Even without too much knowledge of higher mathematics, you may find that these are all numbers smaller than 11. Know-it-all info Note the small but subtle difference between the characters ≤ and < (or ≥ and > ): While it is still allowed if the associated expressions ≤ are equal, it is forbidden with < . In other words, the statement 42 ≤ 42 is completely correct, whereas 42 < 42 is simply wrong. Since you probably don’t want to discuss four different forms of inequality any more than I do, I will concentrate on one of them in the following, based on the following consideration: • Although the meaning of the signs ≤ and < as just emphasized is not the same, the methods for transforming and thus solving the inequalities associated with them are exactly the same; therefore in the following I will limit myself to the real inequalities, that is, expressions associated with < or > .

2.2 What Can You Do With an Inequality Without Changing Its Solution Set?

35

• Expression1 < Expression2 obviously means exactly the same as Expression2 > Expression1 ; therefore I will only consider with < related expressions in the following. In the following we will deal with the solution of inequalities containing an unknown. What exactly is meant by “solving,” I wrote down at the beginning of the previous chapter of the equations; you may want to turn back to it, I will wait here for you.

2.2

What Can You Do With an Inequality Without Changing Its Solution Set?

Admittedly, I used copy and paste to get the title from the first chapter, one has to work efficiently nowadays. However, with the following rule about allowed manipulations of inequalities, I can’t do that without further ado, but I have to pay hellish attention; and you should do the same when reading. Rule The following manipulations of an inequality do not change its solution set and may therefore be used as often as desired: • Adding or subtracting the same number on both sides of the inequality • Multiplying both sides of the inequality with the same positive number • Dividing both sides of the inequality by the same positive number Did you find the difference to the equations? Well, maybe it wasn’t that difficult, because I wrote it in italics: While in equations, multiply and divide by any number except zero, here it’s only allowed with positive numbers. And that’s a good thing: For example, the inequality −1 < 2

(2.3)

is absolutely correct, because −1 is further to the left on the number line than 2. But if I multiply it with the negative number −3, it becomes 3 < −6

36

2

Inequalities

which is certainly not right. But what if for some reason you are forced to multiply both sides of an inequality with a negative number? Well, with a little extra effort, you can handle that too. I pedantically write this down again as a rule; this is to be understood as a supplement to the previous one. Rule The following manipulations of an inequality do not change its solution set and may therefore be used as often as desired: • Multiplying both sides of the inequality with the same negative number and exchanging both sides at the same time • Dividing both sides of the inequality by the same negative number with simultaneous permutation of both sides If I observe this new rule and again multiply the inequality (2.3) with −3, it becomes −6 < 3, which has a much better truth value than the false result determined above. Chitchat In many other books you will find at the end of the two points of the rule just formulated the phrase, “with simultaneous inversion of the inequality sign”; in this case the solution of (2.3) would be 3 > −6. This is of course just as correct as—in all modesty—my formulation above, I have chosen it because I want to concentrate on the “ < ” sign as written above; and I can sometimes be quite stubborn, ask my wife.

2.3

Linear Inequalities

Just as with equations, there are also some types of inequalities which occur again and again and for which there are therefore “ready-made” solution formulas. In this text I will concentrate mainly on linear inequalities and modifications of them; at the end of the chapter I will also tell you why. For once, I don’t want to give an example—I get older—but rather give the definition right away and explain it with examples afterward.

2.3 Linear Inequalities

37

Definition 2.2 An inequality of form ax < b with real numbers a and b, where a must not be zero, is called linear inequality. One asks here, for which values of x the product ax is smaller than the given number b. First examples of this are 3x < 42

(2.4)

−2x < 12.

(2.5)

or perhaps

But what is the answer to this question? Well, in the last section I wrote that you can divide both sides of an inequality by the same positive number without changing its solution set. This is practical, because 3 is certainly a positive number, and therefore I am allowed to divide both sides of (2.4) by 3 and thus get rid of the factor before x. (By the way, I may divide both sides of (2.5) as well as any other inequality by 3, but that would not be very useful. Not funny? OK, forget it.) After dividing by 3, (2.4) becomes the expression x < 14, and this is already the solution of this inequality; in words: All numbers smaller 371 than 14 solve (2.4), for example 13, 372 , 0, −1, and −9999. I cannot list all solutions here, because there are infinitely many, and this is quite typical for inequalities: In contrast to equations, they usually do not have finitely many solutions. Know-it-all info If you want it to be formulated in a mathematically correct way, you have to describe the solution set L of (2.4) like this L = {x|x < 14}. So you will find this in many mathematics books—should you ever pick one up again—but I will hardly ever use it again.

38

2

Inequalities

Let us now finally take a look at the solution of (2.5). Here, of course, it makes sense to divide by −2 to get rid of the pre-factor of x. I do that too, but I have to keep in mind that −2 is negative, and thus the extended rule about swapping sides at the same time comes into play. To cut a long story short: (2.5) becomes −6 < x. Solution of this inequality are all numbers x, which are larger than −6, that is, which are further to the right on the number line than this number. This applies, for example, to x = −5; set this value to the left side of (2.5) for checking purposes. Following the good old rule “minus times minus equals plus” you will get (−2) · (−5) = 10, and 10 is certainly less than 12. What I have just done with the special numbers 3 and −2, I can also do with general positive or negative pre-factors a of the inequality and thus obtain the following rule for their solution: Rule The solution of the linear inequality ax < b consists of all numbers x to which applies: x
0, a

or b < x, if a < 0. a We have already looked at two examples of this before (which is typical mathematical English, because you won’t get the solution by looking at it alone), so I will now just add one more. I will also show you that the sign of b, the right side, does not matter at all. b can be as negative as it wants to be, it does not change the rule.

Example 2.2

The linear inequality to be solved is −3x < −12.

(2.6)

2.4 Linear Inequalities That Have Yet to be Sorted

39

True to the rule, I now have to divide −12 by −3—that makes 4—and swap both sides. I get the solution 4 < x. All numbers x, which are greater than 4, thus fulfill the inequality (2.6). As a rather arbitrary example, I set x = 10 and get −30 < −12; and that is correct. You hesitate? Well, when you have negative numbers, you have to keep in mind that the smaller of two is the one that is greater in amount. Unusual, but correct. Or perhaps the number line that has already been strained several times will help you: Of two negative numbers, the one that is further to the left is smaller; and that is certainly the case for −30 compared to −12.

2.4

Linear Inequalities That Have Yet to be Sorted

With this somewhat cryptic headline I mean the fact that linear inequalities will seldom appear in the same beautifully sorted form ax < b. Most of the time you have to get them into this form by separating the parts with x and the constants. Still pretty cryptic, I admit. I’d rather make an example. Example 2.3

The inequality to be solved is 4x − 3 < 2(x − 2) + 5.

(2.7)

First, I undo the parenthesis on the right side and summarize the constants here as well; this gives 4x − 3 < 2x + 1. Now I subtract 2x on both sides of the inequality and add 3 at the same time. This gives 2x < 4, thus a linear inequality in standard form. Its solution is x < 2, and this is therefore also the solution of the initial Eq. (2.7). Actually, this is always the method, if in an inequality x only occurs linearly, that is, without a superscript, and also not as input of a function,

40

2

Inequalities

and also not multiplied by itself: You multiply if necessary, summarize and then separate all expressions with x from the constants like hazardous waste. There is no rigid rule for this, I prefer to illustrate the procedure again with an example.

Example 2.4

The inequality to be solved is −7x + 3 + 5(2x − 1) < −(x + 3) + 5, whose practical relevance I do not want to discuss here; it is just another example. Exceptionally without disturbing intermediate comments, I give the transformation steps in the following: −7x + 3 + 10x − 5 < −x − 3 + 5 3x − 2 < −x + 2 4x < 4 I leave the solution of this to you in my confidence. Little tip: You shouldn’t land very far from x < 1. And that’s it with this small section, because I think I could already make the procedure clear with two examples and a few additional remarks.

2.5

Linear Inequalities That Are Not Immediately Obvious

Just like equations—there, under the same heading, I had treated the quadratic ones exemplarily—there are also inequalities, which are not seen as having the same simple structure. I will show you some examples. Example 2.5

The inequality (x − 1) · (1 + 2x) < x · (2x + 5) + 5

(2.8)

2.5 Linear Inequalities That Are Not Immediately Obvious

41

does not look very linear at first glance, but we will see that this is deceptive. To simplify the inequality, I first multiply both sides out; this leads to x − 1 + 2x 2 − 2x < 2x 2 + 5x + 5 Now I can subtract 2x 2 on both sides, which makes this square expression disappear completely; it just remains −x − 1 < 5x + 5 or −6x < 6. The solution to this is (Attention, check learning objectives!) −1 < x, that is, all numbers greater than −1. In order to at least make it plausible to you that we are not making nonsense here, I would like to insert two numbers from the solution set into the initial equation and see whether they give a true result. Of course, this is not a complete sample, but it is a test, after all. The simplest number x that I can think of for the condition −1 < x is x = 0. If I insert this value in (2.8), it becomes −1 < 5, so a true statement. The next number that comes to my mind is x = 42 (mathematicians are strange people). This value makes (2.8) the inequality 3485 < 3743. This is obviously correct as well.

42

2

Inequalities

Example 2.6

The second and last example of this small section deals with the inequality x < (x 3 + x 2 + x + 1) · (1 − x) + x 4 .

(2.9)

Doesn’t look very linear either, right? I can understand that, but it’s misleading. If you multiply the right side of the figure, you get (x 3 + x 2 + x + 1) − (x 4 + x 3 + x 2 + x) + x 4 . Here, however, pretty much everything is removed, and only a simple 1 remains; inequality (2.9) is therefore simply x < 1, and at the same time is already solved. So you see: Even if an inequality does not look like a linear one at first, one should not despair and first multiply, summarize, and simplify everything that is possible. Maybe you will be lucky and everything non-linear will stand out.

2.6

Fraction Inequalities

I have saved the most interesting type of inequality I want to introduce to you until the end: Inequalities in which the unknown (also) occurs in the denominator of a fraction. Example 2.7

A first example of this is 3x + 2 < 4. x

(2.10)

Maybe you are thinking: What is the problem, you multiply both sides by x, then the denominator on the left side disappears and you have a beautiful linear inequality in front of you? In principle that is correct, BUT:

2.6 Fraction Inequalities

43

Attention! Attention! Attention! Since one does not yet know whether the denominator, that is, x, is positive or negative, and since it again depends on whether or not one has to swap both sides of the inequality after multiplication, one has to distinguish between these two cases and treat them separately. And this is exactly what I am doing now. Case 1: 0 < x. In this case I can multiply the inequality (2.10) without changing anything and get 3x + 2 < 4x, so the condition 2 < x. Now toward the end of this booklet you have to concentrate well once again: In this first case, which we are currently investigating, x must therefore fulfil two conditions: It must be greater than 0 due to the case distinction, and it must be greater than 2 as a result of the transformation. So, in general, these are two conditions that will contribute to the formulation of the solution, but in this example—and this will very often be the case—the second condition is sharper than the first one (since any number greater than 2 is automatically greater than 0), so that in reality only one condition remains here, namely 2 < x.

(2.11)

Case 2: x < 0. In this case, after multiplying, I have to swap the two sides and get 4x < 3x + 2, so x < 2. Here now a similar phenomenon occurs as in the first case: The solutions must meet the two conditions x < 0 and x < 2, since the first undoubtedly implies the second, the whole thing is reduced in this second case to x < 0.

(2.12)

44

2

Inequalities

The total solution set of the inequality (2.10) thus consists of all numbers x which satisfy either (2.11) or (2.12), that is, which are either greater than two or less than zero. Of course it can happen that the denominator does not only contain an isolated x; here is an example of this, too.

Example 2.8

The inequality to be solved is x −1 < 1. 2x + 1

(2.13)

Again, the two sign situations of the denominator must be distinguished in two separate cases. Case 1: 0 < 2x + 1, so − 21 < x. In this case I can multiply the inequality (2.13) and obtain x − 1 < 2x + 1, so −2 < x. So the unknown x must fulfill two conditions here too; it must be greater than − 21 due to the case distinction and it must be greater than −2 as a result of the transformation. And just as in the first example, one of the two conditions— here the first one—is sharper than the other (since every number greater than − 21 is automatically greater than −2), which is why also here in reality only one condition remains, namely −

1 < x. 2

(2.14)

Case 2: 2x + 1 < 0, so x < − 21 . Here too I multiply the inequality (2.13) by, but now I have to reverse the two sides and obtain 2x + 1 < x − 1,

2.6 Fraction Inequalities

45

so x < −2. As in case 1, here too we have two conditions on x, namely x < − 21 and x < −2, one of which automatically entails the other. In case 2, the solution is therefore x < −2.

(2.15)

There are no more cases, since 2x + 1 certainly must not be zero, so that I can extract from (2.14) and (2.15) the solution set of the inequality (2.13); it consists of all numbers which are either greater than −2 or less than − 21 . If you would like to have another formal language version at the end of this text: The solution set of (2.15) is 

 1 L = x ∈ I R|x < −2 or − < x . 2  A third and—to your consolation—last example should help you to practice the procedure and at the same time show that a solution does not exist in every case. Example 2.9

This last example concerns the inequality. 2x − 6 < 0. x −1

(2.16)

Here too, of course, I must distinguish between two cases: Case 1: 0 < x −1, so 1 < x. In this case, I can multiply by the denominator without swapping the sides of the inequality and obtain 2x − 6 < 0 (since the right hand side is multiplied by zero) x < 3.

(2.17)

There are two conditions: 1 < x and x < 3. This time, neither of them implies the other, so both must be taken into account. The solution set thus consists of all numbers x that are simultaneously greater than 1 and less than 3, that is, that lie between 1 and 3 on the number line. Case 2: x − 1 < 0, so x < 1. Here too I can multiply by the denominator, but now I have to swap the sides of the inequality 0 < 2x − 6 and obtain

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2

Inequalities

6 < 2x or 3 < x.

(2.18)

A possible solution x of the inequality would thus have to be both smaller than 1 and larger than 3. However, such a number has not been invented until today, therefore in this case there is no solution of the inequality (2.16), and the only solutions are those which resulted in case 1. I consider it didactically appropriate, at the end of this book, to give you the rule for solving such fractional inequalities only after the examples; when you have read it, perhaps you can work through the three examples of this section again after that. Rule To solve a linear fractional inequality of the form ax + b < e, cx + d where a, b, c, d, and e are numbers, and c and d must not be zero at the same time, one proceeds as follows • A distinction is made between the two cases 0 < cx + d and cx + d < 0. • Case 1: 0 < cx +d. Multiply by the denominator and you get the linear inequality

ax + b < ecx + ed, which can be solved using the methods described above. Then the solution condition obtained in this way must be reconciled with the condition 0 < cx + d. Depending on the situation, the following cases can occur: • One of the two conditions implies the other; then only the stricter condition needs to be taken into account to formulate the solution. • Both conditions can be fulfilled at the same time and neither implies the other; then both conditions must be taken into account when formulating the solution. • The two conditions are contradictory, that is, if one is fulfilled, the other is violated; then there is no solution in case 1.

2.6 Fraction Inequalities

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– Case 2: cx + d < 0. Multiply by the denominator and you get the linear inequality

ecx + ed < ax + b. Know-it-all info No, actually not an info, but more an invitation to think, because in the meantime you are the know-it-all: Why must c and d not be zero at the same time? This brings us to the end of this booklet. That wasn’t so bad, was it? By the way, if you were waiting for me to introduce you to quadratic inequalities, I have to disappoint you: Quadratic inequalities can be quite violent, and since this text, according to its subtitle, is (also) addressed to non-mathematicians, I wanted to spare you this. We have already achieved enough together.

What You Learned From This essential

• Linear equations always have exactly one solution, which can easily be determined • There are also ready-made solution formulas for quadratic equations • Often complicated looking equations, for example root or fractional equations, can be traced back to quadratic or even linear equations and thus be solved quite easily • Solving inequalities is not as difficult as one often believes

© Springer Fachmedien Wiesbaden GmbH, part of Springer Nature 2021 G. Walz, Equations and Inequalities, Springer essentials, https://doi.org/10.1007/978-3-658-32720-0

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References

Kemnitz, A. (2014). Mathematik zum Studienbeginn (11th ed.). Heidelberg: SpringerSpektrum. Papula, L. (2014). Mathematik für Ingenieure und Naturwissenschaftler (Vol. 3). Wiesbaden: Vieweg + Teubner. Rießinger, T. (2017). Mathematik für Ingenieure (10th ed.). Heidelberg: Springer. Stingl, P. (2009). Mathematik für Fachhochschulen (8th ed.). München: Hanser Fachbuch. Stingl, P. (2013). Einstieg in die Mathematik für Fachhochschulen (5th ed.). München: Hanser Fachbuch. Walz, G. (2020). Mathematik für Hochschule und duales Studium (3rd ed.). Heidelberg: Springer-Spektrum. Walz, G., Zeilfelder, F., & Rießinger, T. (2019). Brückenkurs Mathematik (5th ed.). Heidelberg: Springer-Spektrum.

© Springer Fachmedien Wiesbaden GmbH, part of Springer Nature 2021 G. Walz, Equations and Inequalities, Springer essentials, https://doi.org/10.1007/978-3-658-32720-0

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