BBA120 Business Mathematics - University of Lusaka · BBA 120 Business Mathematics 3 Unit 2:...

SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT

BBA120 BUSINESS MATHEMATICS / MATHEMATICAL ANALYSIS

Unit Two

Moses Mwale

e-mail: [email protected]

BBA 120 Business Mathematics

Contents

Unit 2: Equations and Inequalities 3

2.0 Arithmetic Operations ................................................................................................. 3 2.1. Addition and subtraction .................................................................................. 3

Examples ............................................................................................... 4 2.2 Multiplication and Division............................................................................... 4

Examples ............................................................................................... 5 2.3 Order of Operations ........................................................................................... 5

Examples ............................................................................................... 5

2.4 Fractions ............................................................................................................ 6 2.4.1 Addition and Subtraction of Fractions .................................................. 7

Example ................................................................................................. 7 2.4.2 Multiplication and Division of fractions ............................................... 7

Examples ............................................................................................... 8 2.5 Decimal Representation of Numbers ................................................................ 8

2.5.1 Converting Decimals to Fractions ........................................................ 9 2.5.2 Converting repeating decimals to fractions .......................................... 9

Example #1: ......................................................................................... 10 Example #2: ......................................................................................... 11

2.5.2 Significant Figures .............................................................................. 12

Example ....................................................................................................... 13 2.1 Linear Equations ....................................................................................................... 13

2.2 Simple Inequalities ................................................................................................... 15 2.2.1 Solving Inequalities ...................................................................................... 16

Rational Inequalities .................................................................................... 19 Exercise ............................................................................................... 20

2.3 Percentages ............................................................................................................... 21 2.4 Powers and Indices ................................................................................................... 25 2.5 Quadratic Equations .................................................................................................. 28

2.5.1 Solving Quadratic Equations ........................................................................ 29

Solving by Factorization Method................................................................. 29 The Quadratic Formula ................................................................................ 31 Completing the Square Method ................................................................... 34 Solutions of a Quadratic Equation ............................................................... 35

2.6 Absolute Value ......................................................................................................... 35

Solving Absolute Value Equations .............................................................. 36 Practice Problems ................................................................................ 37


3

Unit 2: Equations and Inequalities

2.0 Arithmetic Operations

Arithmetic or arithmetic’s is

the oldest and most elementary

branch of mathematics, used by

almost everyone, for tasks

ranging from simple day-to-day

counting to advanced science

and business calculations. It

involves the study of quantity,

especially as the result of

operations that combine

numbers. In common usage, it refers to the simpler properties when using

the traditional operations of addition, subtraction, multiplication and

division with smaller values of numbers.

2.1. Addition and subtraction

Adding: If all the signs are the same, simply add all the terms and give

the answer with the common overall sign.

Subtracting: When subtracting any two numbers or two similar terms,

give the answer with the sign of the largest number or term.

If a and b are any two numbers, then we have the following rules

𝑎 + (−𝑏) = 𝑎 − 𝑏,

𝑎 − (+𝑏) = 𝑎 − 𝑏,

𝑎 − (−𝑏) = 𝑎 + 𝑏.

Thus we can regard −(−𝑏) as equal to +𝑏.

We consider a few examples:

4 + (−1) = 4 − 1 = 3,

and

3 − (−2) = 3 + 2 = 5.

The last example makes sense if we regard 3 − (−2) as the difference

between 3 and −2 on the number line.

Note that a − b will be negative if and only if a < b. For example,

−2 − (−1) = −2 + 1 = −1 < 0.

4 Unit 2: Equations and Inequalities

Remember

It is useful to remember

that a minus sign is a -1,

so -5 is read as −1 × 5

Also

0 × (any real number) = 0

0 ÷ (any real number) = 0

But you cannot divide by

0

Brackets are used for

grouping terms together

in maths for:

(i) Clarity

(ii) Indicating the order in

which a series of

operations should be

carried out

Examples

Add/subtract with numbers, mostly

5 + 8 + 3 = 16

5 + 8 + 3 + 𝑣 = 16 + 𝑦

(The y-term is different, so it cannot be

added to the others)

Add/subtract with variable terms

5𝑥 + 8𝑥 + 3𝑥 = 16𝑥

5𝑥 + 8𝑥 + 3𝑥 + 𝑣 = 16𝑥 + 𝑦

5𝑥𝑣 + 8𝑥𝑣 + 3𝑥𝑣 + 𝑣 = 16𝑥𝑣 + 𝑦

2.2 Multiplication and Division

Multiplication

If a and b are any two positive numbers, then we have the following rules

for multiplying positive and negative numbers:

𝑎 × (−𝑏) = −(𝑎 × 𝑏),

(−𝑎) × 𝑏 = −(𝑎 × 𝑏),

(−𝑎) × (−𝑏) = 𝑎 × 𝑏.

So multiplication of two numbers of the same sign gives a positive

number, while multiplication of two numbers of different signs gives a

negative number.

For example, to calculate 2 × (−5), we multiply 2 by 5 and then place a

minus sign before the answer. Thus, 2 × (−5) = −10.

These multiplication rules give, for example,

(−2) × (−3) = 6, (−4) × 5 = −20, 7 × (−5) = −35.

Division

The same rules hold for division because it is the same sort of operation

as multiplication, since

𝑎

𝑏= 𝑎 ×

1

𝑏

So the division of a number by another of the same sign gives a positive

number, while division of a number by another of the opposite sign gives

a negative number.


5

Examples

a) 5 × 7 = 35

b) −5 × −7 = 35

c) −5 × 7 = −35

d) (−15) ÷ (−3) = 5

e) (−16) ÷ 2 = −8

f) 2 ÷ (−4) = −1

2

g) (−𝑥)(−𝑦) = 𝑥𝑦

h) 2(𝑥 + 4) = 2𝑥 + 8

i) (𝑥 − 3)(𝑥 + 4) = 𝑥(𝑥 + 4) − 3(𝑥 + 4)

= 𝑥2 + 4𝑥 − 3𝑥 − 12

= 𝑥2 + 𝑥 − 12

2.3 Order of Operations

The order in which operations in an arithmetical expression are

performed is important. Consider the calculation

12 + 8 ÷ 4.

Different answers are obtained depending on the order in which the

operations are executed.

If we first add together 12 and 8 and then divide by 4, the result is 5.

However, if we first divide 8 by 4 to give 2 and then add this to 12, the

result is 14.

Therefore, the order in which the mathematical operations are performed

is important and the convention is as follows:

Brackets, exponents, division, multiplication, addition, and subtraction.

This convention has the acronym BEDMAS.

However, the main point to remember is that if you want a calculation to

be done in a particular order, you should use brackets to avoid any

ambiguity.

Examples

Evaluate the expressions

a) 3 ∗ ( 5 + 8 ) − 22 ÷ 4 + 3

Brackets or Parenthesis first: 5 + 8 = 13

3 ∗ 13 − 22 ÷ 4 + 3

Note: Other texts

use the Acronyms

BODMAS or

PEMDAS.


Exponent next: square the 2 or 22 = 4

3 ∗ 13 − 4 ÷ 4 + 3

Multiplication and Division next (3 ∗ 13) (4 ÷ 4)

left to right:

39 − 1 + 3

Addition and Subtraction next

left to right:

39 − 1 + 3 = 41

b) 23 × 3 + (5 − 1).

c) 4 – 3[4 – 2(6 – 3)] ÷ 2.

2.4 Fractions

A fraction is a number that expresses part of a whole. It takes the form 𝑎

𝑏

where a and b are any integers except that b ≠ 0.

The integers a and b are known as the numerator and denominator of

the fraction, respectively.

Examples of statements that use fractions are,

3

5 of students in a lecture may be female or

1

3 of a person’s income may be taxed by the government.

Fractions may be simplified to obtain what is known as a reduced

fraction or a fraction in its lowest terms. This is achieved by identifying

any common factors in the numerator and denominator and then

cancelling those factors by dividing both numerator and denominator by

them.

For example, consider the simplification of the fraction 27

45. Both the

numerator and denominator have 9 as a common factor since 27 = 9 × 3

and 45 = 9 × 5 and therefore it can be cancelled:

27

45=

3 × 9

9 × 5=

3

5

We say that 27

45 and

3

5 are equivalent fractions and that

3

5 is a reduced

fraction.

Comparing two fractions

To compare the relative sizes of two fractions and also to add or subtract

two fractions, we express them in terms of a common denominator.

The common denominator is a number that each of the denominators of

the respective fractions divides, i.e., each is a factor of the common

denominator.

Note that in the

fraction 𝑎

𝑏

a can be greater

than b


7

Suppose we wish to determine which is the greater of the two fractions 4

9 and

5

11.

The common denominator is 9 × 11 = 99. Each of the denominators (9

and 11) of the two fractions divides 99.

We follow a similar procedure when we want to add or subtract two

fractions.

2.4.1 Addition and Subtraction of Fractions

To add or subtract fractions we use the following method;

Step 1: Take a common denominator, that is, a number or term which is

divisible by the denominator of each fraction to be added or subtracted. A

safe bet is to use the product of all the individual denominators as the

common denominator.

Step 2: For each fraction, divide each denominator into the common

denominator, then multiply the answer by the numerator.

Step 3: Simplify your answer if possible.

Example

Simplify

a) 13

24−

5

16

b) 1

7+

2

3−

4

5

c) 𝑥

7+

2𝑥

3−

4𝑥

5

2.4.2 Multiplication and Division of fractions

To multiply together two fractions, we simply multiply the numerators

together and multiply the denominators together:

𝑎

𝑏×

c

d=

a × b

c × b=

ac

bd

To divide one fraction by another, we multiply by the reciprocal of the

divisor where the reciprocal of the fraction a/b is defined to be 𝑏

𝑎 provided

a, b ≠ 0.

That is

𝑎

𝑏÷

𝑐

𝑑=

a × d

c × b=

𝑎𝑑

𝑐𝑏

Least Common

Multiple (LCM) The least common

multiple (also called

the lowest common

multiple or smallest

common multiple) of

two integers a and b,

usually denoted by

LCM(a, b), is the

smallest positive

integer that is divisible

by both a and b.

If either a or b is 0,

LCM(a, b) is defined

to be zero.

For example the LCM

of 4 and 6 is 12

Greatest Common

Divisor (gcd)

The greatest common

divisor (gcd), also

known as the greatest

common factor (gcf),

or highest common

factor (hcf), of two or

more non-zero

integers, is the largest

positive integer that

divides the numbers

without a remainder.

For example, the GCD

of 8 and 12 is 4.


Examples

a) (2

3) (

5

7) =

(2)(5)

(3)(7)=

10

21

b) (−2

3) (

7

5) =

(−2)(7)

(3)(5)= −

14

15

c) 3 ×2

5= (

3

1) (

2

5) =

(3)(2)

(1)(5)=

6

5= 1

1

5

d) (3

𝑥)

(𝑥+3)

(𝑥−5)=

3(𝑥+3)

𝑥(𝑥−5)=

3𝑥+9

𝑥2−5𝑥

e) (

2

3)

(5

11)

= (2

3) (

11

5) =

22

15

f)

2𝑥

𝑥+𝑦3𝑥

2(𝑥−𝑦)

=2𝑥

𝑥+𝑦 2(𝑥−𝑦)

3𝑥

=4𝑥(𝑥−𝑦)

3𝑥(𝑥+𝑦)=

4(𝑥−𝑦)

3(𝑥+𝑦)

2.5 Decimal Representation of Numbers

A fraction or rational number may be converted to its equivalent decimal

representation by dividing the numerator by the denominator.

For example, the decimal representation of 3

4 is found by dividing 3 by 4

to give 0.75. This is an example of a terminating decimal since it ends

after a finite number of digits.

The following are examples of rational numbers that have a terminating

decimal representation:


9

1

8= 0.125, and

3

25= 0.12.

Some fractions do not possess a finite decimal representation – they go

on forever. The fraction 1

3 is one such example. Its decimal representation

is 0.3333... where the dots denote that the 3’s are repeated and we write

1

3= 0. 3̇

where the dot over the number indicates that it is repeated indefinitely.

This is an example of a recurring decimal.

All rational numbers have a decimal representation that either terminates

or contains an infinitely repeated finite sequence of numbers. Another

example of a recurring decimal is the decimal representation of 1/13:

1

13= 0.0769230769230 … = 0.07̇69230̇

where the dots indicate the first and last digits in the repeated sequence.

2.5.1 Converting Decimals to Fractions

To convert a decimal to a fraction, you simply have to remember that the

first digit after the decimal point is a tenth, the second a hundredth, and

so on.

For example,

0.2 =2

10=

1

5

and

0.375 =375

1000=

3

8

2.5.2 Converting repeating decimals to fractions

When converting repeating decimals to fractions, just follow the two

steps below carefully.

Step 1: Let x equal the repeating decimal you are trying to convert to a

fraction

Step 2: Examine the repeating decimal to find the repeating digit(s)

Step 3: Place the repeating digit(s) to the left of the decimal point

Step 4: Place the repeating digit(s) to the right of the decimal point

In some text

books, a bar

notation is used

instead of a dot

e.g.

52.71656565…

(65 repeating

infinitely often)

may be written

52.7165̅̅̅̅


Step 5: Subtract the left sides of the two equations. Then, subtract the

right sides of the two equations

As you subtract, just make sure that the difference is positive for both

sides

Now let's practice converting repeating decimals to fractions with two

good examples

Example #1:

What rational number or fraction is equal to 0.55555555555

Step 1: x = 0.5555555555

Step 2: After examination, the repeating digit is 5

Step 3: To place the repeating digit ( 5 ) to the left of the decimal point,

you need to move the decimal point 1 place to the right

Technically, moving a decimal point one place to the right is done by

multiplying the decimal number by 10.

When you multiply one side by a number, you have to multiply the other

side by the same number to keep the equation balanced

Thus, 10x = 5.555555555

Step 4: Place the repeating digit(s) to the right of the decimal point

Look at the equation in step 1 again. In this example, the repeating digit

is already to the right, so there is nothing else to do.

x = 0.5555555555

Step 5: Your two equations are:

10x = 5.555555555

x = 0.5555555555

10x - x = 5.555555555 − 0.555555555555

9x = 5

Divide both sides by 9

x = 5/9


11

Example #2:

What rational number or fraction is equal to 1.04242424242

Step 1: x = 1.04242424242

Step 2:After examination, the repeating digit is 42

Step 3:To place the repeating digit ( 42 ) to the left of the decimal point,

you need to move the decimal point 3 place to the right

Again, moving a decimal point three place to the right is done by

multiplying the decimal number by 1000.

When you multiply one side by a number, you have to multiply the other

side by the same number to keep the equation balanced

Thus, 1000x = 1042.42424242

Step 4:Place the repeating digit(s) to the right of the decimal point

In this example, the repeating digit is not immediately to the right of the

decimal point.

Look at the equation in step 1 one more time and you will see that there is

a zero between the repeating digit and the decimal point

To accomplish this, you have to move the decimal point 1 place to the

right

This is done by multiplying both sides by 10

10x = 10.4242424242

Step 5:Your two equations are:


1000x = 1042.42424242

10x = 10.42424242

1000x - 10x = 1042.42424242 − 10.42424242

990x = 1032

Divide both sides by 990:

x = 1032/990

To master this lesson about converting repeating decimals to fractions,

you will need to study th2e two examples above carefully and practice

with other examples

2.5.2 Significant Figures

Sometimes we are asked to express a number correct to a certain number

of decimal places or a certain number of significant figures.

Suppose that we wish to write the number 23.541638 correct to two

decimal places. To do this, we truncate the part of the number following

the second digit after the decimal point:

23.54 | 1638.

Then, since the first neglected digit, 1 in this case, lies between 0 and 4,

then the truncated number, 23.54, is the required answer.

If we wish to write the same number correct to three decimal places, the

truncated number is

23.541 | 638,

and since the first neglected digit, 6 in this case, lies between 5 and 9,

then the last digit in the truncated number is rounded up by 1.

Therefore, the number 23.541638 is 23.542 correct to three decimal

places or, for short, ‘to three decimal places’.

To express a number to a certain number of significant figures, we

employ the same rounding strategy used to express numbers to a certain

number of decimal places but we start counting from the first non-zero

digit rather than from the first digit after the decimal point.

For example,

72,648 = 70,000 (correct to 1 significant figure)

= 73,000 (correct to 2 significant figures)

= 72,600 (correct to 3 significant figures)


13

= 72,650 (correct to 4 significant figures),

and

0.004286 = 0.004 (correct to 1 significant figure)

= 0.0043 (correct to 2 significant figures)

= 0.00429 (correct to 3 significant figures).

Note that 497 = 500 correct to 1 significant figure and also correct to 2

significant figures.

Example

a) A teacher may ask for an exact answer to the problem “What

is the length of the diagonal of a square whose sides have

length 2?” The exact answer is √8. An approximate answer

is 2.8284.

b) A somewhat more complicated example is 3227/555 =

5.8144144144…, here the decimal representation becomes

periodic at the second digit after the decimal point, repeating

the sequence of digits "144" indefinitely.

2.1 Linear Equations

A mathematical statement setting two algebraic expressions equal to each

other is called an equation.

The simplest type of equation is the linear equation in a single variable

or unknown, which we will denote by x for the moment.

In a linear equation, the unknown x only occurs raised to the power 1.

The following are examples of linear equations:

i. 5𝑥 + 3 = 11,

ii. 1 − 4𝑥 = 3𝑥 + 7,

A linear equation may be solved by rearranging it so that all terms

involving x appear on one side of the equation and all the constant terms

appear on the other side.

Not all equations have solutions. In fact, equations may have no solutions

at all or may have infinitely many solutions. Each of these situations is

demonstrated in the following examples.

Case 1: Unique solutions: An example of this is given above: 𝑥 + 4 =

10 etc.

The solution of an

equation is simply

the value or values

of the unknown(s)

for which the left-

hand side (LHS) of

the equation is

equal to the right-

hand side (RHS).


Case 2: Infinitely many solutions: The equation, 𝑥 + 𝑦 = 10 has

solutions (𝑥 = 5, 𝑦 = 5); (𝑥 = 4, 𝑦 = 6); (𝑥 = 3, 𝑦 = 7), etc. In

fact, this equation has infinitely many solutions or pairs of values (x, y)

which satisfy the formula, 𝑥 + 𝑦 = 10.

Case 3: No solution: The equation, 0(𝑥) = 5 has no solution. There is

simply no value of x which can be multiplied by 0 to give 5.

Solving equations can involve a variety of techniques, many of which

will be covered later

Examples

(a) Given the equation 𝑥 + 3 = 2𝑥 − 6 + 5𝑥, solve for x.

Solution

𝑥 + 3 = 2𝑥 − 6 + 5𝑥

𝑥 + 3 = 7𝑥 − 6 (adding the x terms on the RHS)

𝑥 + 3 + 6 = 7𝑥 (bringing over –6 to the other side)

𝑥 + 9 = 7𝑥

9 = 7𝑥 − 𝑥 (bringing x over to the other side)

9 = 6𝑥

9

6 = 𝑥 (dividing both sides by 6)

1.5 = 𝑥

(b) Solve the equation 𝑥

4− 3 =

𝑥

5+ 1

Solution.

We go through the solution step-by-step. The idea is to rearrange

the equation so that all terms involving x appear on the left-hand

side and all the constant terms appear on the right-hand side.

Once this is done, the terms involving fractions are simplified.

1. Subtract 𝑥

5 from both sides:

𝑥

4−

𝑥

5− 3 = 1

2. Add 3 to both sides

𝑥

4−

𝑥

5= 1 + 3 = 4

3. Simplify the left-hand side by expressing it as a single fraction.

This is achieved by expressing each of the fractions in terms of

their lowest common denominator, 20. In the case of the first

fraction, both the numerator and denominator are multiplied by 5,

and in the case of the second fraction they are both multiplied by

4, i.e.,

𝑥

4=

5𝑥

5×4=

5𝑥

20 and

𝑥

5=

4𝑥

4×5 =

4𝑥

20

A common form of a linear equation in the two variables x

and y is

𝒚 = 𝒎𝒙 + 𝒄

where m and c designate

constants. The origin of the name

"linear" comes from the fact that

the set of solutions of such an equation

forms a straight line in the plane. In

this particular equation, the constant m

determines the slope or gradient of

that line, and the constant term c determines the


15

Therefore

5𝑥

20=

4𝑥

20= 4

5𝑥 − 4𝑥

20= 4

𝑥

20= 4

4. Finally multiply both sides by 20:

𝑥 = 80.

The solution to this equation is x = 80. Again we can check that this is the

correct solution by substituting x = 80 into the left- and right-hand sides

of our main equation.

2.2 Simple Inequalities

An equation is an equality. It states that the expression on the LHS of the

'=' sign is equal to the expression on the RHS.

An inequality is a relation that holds between two values when they are

different. In an inequality, expression on the LHS is either greater than

(denoted by the symbol >) or less than (denoted by the symbol <) the

expression on the RHS.

For example, 5 = 5 or 5𝑥 = 5𝑥 are equations, while

5 > 3 and 5𝑥 > 3𝑥 are inequalities which read, '5 is greater than 3';

'5x is greater than 3x' (for any positive value of x).


2.2.1 Solving Inequalities

Solving linear inequalities is very similar to solving linear equations,

except for one small but important detail: you flip the inequality sign

whenever you multiply or divide the inequality by a negative.

When both sides of the inequality are multiplied or divided by negative

numbers or variables, then the statement remains true only when the

direction of the inequality changes: > becomes < and vice versa. For

example, multiply both sides of the inequality, 5 > 3 by -2:

5(−2) > 3(– 2) or – 10 > – 6 is not true

5(−2) < 3(−2), or −10 < −6 is true

We show this with some more examples:

Example 2.2.1

Find the range of values for which the following inequalities are true,

assuming that x > 0. State the solution in words and indicate the solution

on the number line.

a) 10 < 𝑥 − 12 b) −75

𝑥> 15 c) 2𝑥 − 6 ≤ 12 − 4𝑥

Solution

(a) 10 < 𝑥– 12 → 10 + 12 < 𝑥 → 22 < 𝑥 (𝑜𝑟 𝑥 > 22).

The solution states: 22 is less than x or .v is greater than 22. It is

represented by all points on the number line to the right of, but not

including, 22, as shown in the number line below

(b) −75

𝑥> 15

Multiply both sides of the inequality by x. Since x > 0, the direction of

the inequality sign does not change.

−75 > 15𝑥

– 5 > 𝑥 dividing both sides by 15

The solution states: -5 is greater than x or x is less than -5.

x cannot be less than -5 and greater than 0 at the same time. So there is no

solution

22 10 0 -10


17

(c) 2𝑥 – 6 < 12 – 4𝑥

2𝑥 + 4𝑥 − 6 < 12 add 4x to both sides

6𝑥 < 12 + 6 add 6 to both sides

6𝑥 < 18

𝑥 < 3 dividing both sides by 6

Solve x2 – 3x + 2 > 0

First, we have to find the x-intercepts of the associated quadratic

equation, because the intercepts are where 𝑦 = 𝑥2 – 3𝑥 + 2 is equal to

zero.

Graphically, an inequality like this is asking us to find where the graph is

above or below the x-axis. It is simplest to find where it actually crosses

the x-axis, so we'll start there.

Factoring, we get 𝑥2 – 3𝑥 + 2 = (𝑥 – 2)(𝑥 – 1) = 0, so 𝑥 = 1 or

𝑥 = 2. Then the graph crosses the x-axis at 1 and 2, and the number line

is divided into the intervals (negative infinity, 1), (1, 2), and (2, positive

infinity).

Between the x-intercepts, the graph is either above the axis (and thus

positive, or greater than zero), or else below the axis (and thus negative,

or less than zero).

There are two different algebraic ways of checking for this positivity or

negativity on the intervals. I'll show both.

1) Test-point method. The intervals between the x-intercepts are (negative

infinity, 1), (1, 2), and (2, positive infinity). I will pick a point (any point)

inside each interval. I will calculate the value of y at that point. Whatever

the sign on that value is, that is the sign for that entire interval.

For (negative infinity, 1), let's say I choose x = 0; then y = 0 – 0 + 2 = 2,

which is positive. This says that y is positive on the whole interval of

(negative infinity, 1), and this interval is thus part of the solution (since

I'm looking for a "greater than zero" solution).

For the interval (1, 2), I'll pick, say, x = 1.5; then y = (1.5)2 – 3(1.5) + 2 =

2.25 – 4.5 + 2 = 4.25 – 4.5 = –0.25, which is negative. Then y is negative

on this entire interval, and this interval is then not part of the solution.

For the interval (2, positive infinity), I'll pick, say, x = 3; then y = (3)2 –

3(3) + 2 = 9 – 9 + 2 = 2, which is positive, and this interval is then part of

the solution. Then the complete solution for the inequality is x < 1 and x >

2. This solution is stated variously as:


inequality notation

interval, or set, notation

number line with parentheses

(brackets are used

for closed intervals)

number line with open dots

(closed dots are used

for closed intervals)

The particular solution format you use will depend on your text, your

teacher, and your taste. Each format is equally valid.999-2011 All Rights

Reserved

2) Factor method. Factoring, I get 𝑦 = 𝑥2 – 3𝑥 + 2 = (𝑥 – 2)(𝑥 – 1). Now I will consider each of these factors separately.

The factor 𝑥 – 1 is positive for x > 1; similarly, x – 2 is positive for x > 2.

Thinking back to when I first learned about negative numbers, I know

that (plus)×(plus) = (plus), (minus)×(minus) = (plus), and (minus)×(plus)

= (minus). So, to compute the sign on y = x2 – 3x + 2, I only really need

to know the signs on the factors. Then I can apply what I know about

multiplying negatives.

First, I set up a grid,

showing the factors

and the number line.

Now I mark the

intervals where each

factor is positive.

Where the factors

aren't positive, they

must be negative.

Now I multiply up the

columns, to compute

the sign of y on each

interval.


19

Then the solution of x2 – 3x + 2 > 0 are the two intervals with the "plus"

signs:

(negative infinity, 1) and (2, positive infinity).

Rational Inequalities

Solve 𝒙

𝒙 – 𝟑 < 𝟐.

First off, I have to remember that I can't begin solving until I have the

inequality in "= 0" format.

Now I need to convert to a common denominator:

...and then I can simplify: Copyri1999-2011 All Rights Reserved

The two factors are – 𝑥 + 6 and 𝑥 – 3. Note that x cannot equal 3, or

else I would be dividing by zero, which is not allowed. The first factor, –x

+ 6, equals zero when x = 6. The other factor, x – 3, equals zero when x =

3. Now, x cannot actually equal 3, so this endpoint will not be included in

any solution interval (even though this is an "or equal to" inequality), but

I need the value in order to figure out what my intervals are. In this case,

my intervals are (negative infinity, 3), (3, 6], and [6, positive infinity).

Note the use of brackets to indicate that 6 can be included in the solution,

but that 3 cannot.

Using the Test-Point Method, I would pick a point in each interval and

test for the sign on the result. I could use, say, 𝑥 = 0, 𝑥 = 4, and 𝑥 = 7.


Using the Factor Method, I solve each factor: – 𝑥 + 6 > 0 for – 𝑥 >

– 6, or 𝑥 < 6; 𝑥 – 3 > 0 for 𝑥 > 3. Then I do the grid:

...fill in the signs on the factors:

...and solve for the sign on the rational function:

So the solution is all x's in the intervals (negative infinity, 3) and [6,

positive infinity).

Exercise

1. Solve the following inequalities, giving your answer in interval notation.

a) 5𝑦+2

4≤ 2𝑦 − 1

b) −3𝑥 + 1 ≤ −3(𝑥 − 2) + 1

c) 0𝑥 ≤ 0

d) 5(3𝑡+1)

3>

2𝑡−4

6𝑡

e) 2𝑥2 − 19𝑥 − 10 ≤ 0

2. Each month last year, Brittany saved more than $50, but less than $150. If

S represents her total savings for the year, describe S by using inequalities.

3. A student has $360 to spend on a stereo system and some compact discs.

If she buys a stereo that costs $219 and the discs cost $18.95 each, find the

greatest number of discs she can buy.

4. The Davis Company manufactures a product that has a unit selling price of

$20 and a unit cost of $15. If fixed costs are $600,000, determine the least

number of units that must be sold for the company to have a profit.


21

5. Suppose a company offers you a sales position with your choice of two

methods of determining your yearly salary. One method pays $35.000 plus

a bonus of 3% of your yearly sales. The other method pays a straight 5%

commission on your sales. For what yearly sales amount is it better to

choose the first method?

2.3 Percentages

A percentage is a number or ratio as a

fraction of 100. It is often denoted using

the percent sign, “%”, or the abbreviation

“pct.”

Although percentages are usually used to

express numbers between zero and one,

any ratio can be expressed as a

percentage. For instance, 111% is 1.11

and −0.35% is −0.0035.

To convert a fraction to a percentage, we multiply the fraction by 100%.

For example,

3

4=

3

4× 100% = 75% ,

and 3

13=

3

13× 100% = 23.077% (to three decimal places).

To perform the reverse operation and convert a percentage to a fraction,

we divide the number by 100. The resulting fraction may then be

simplified to obtain a reduced fraction.

For example,

45% =45

100=

9

20 ,

where the fraction has been simplified by dividing the numerator and

denominator by 5 since this is a common factor of 45 and 100.

To find the percentage of a quantity, we multiply the quantity by the

number and divide by 100. For example,


25% of 140 is 25

100× 140 = 35

and

4% of 5, 200 is 4

100× 5,200 = 208

If a quantity is increased by a percentage, then that percentage of the

quantity is added to the original. Suppose that an investment of £300

increases in value by 20%. In monetary terms, the investment increases

by

20

100× 300 = £60

and the new value of the investment is

£300 + £60 = £360.

In general, if the percentage increase is r%, then the new value of the

investment comprises the original and the increase. The new value can be

found by multiplying the original value by the factor

1 +𝑟

100 .

It is easy to work in the reverse direction and determine the original value

if the new value and percentage increase is known. In this case, one

simply divides by the factor

1 +𝑟

100 .

Examples

a) An investment rises from $2500 to $3375. Express the increase as a

percentage of the original.

Solution.

The rise in the value of the investment is

3375 − 2500 = 875

As a fraction of the original this is

875

2500= 0.35

This is the same as 35 hundredths, so the percentage rise is 35%.

b) At the beginning of a year, the population of a small village is 8400. If

the annual rise in population is 12%, find the population at the end of the

year.

Solution

As a fraction


23

12% is the same as 12

100 = 0.12

so the rise in population is

0.12 × 8400 = 1008

Hence the final population is

8400 + 1008 = 9408

c) In a sale, all prices are reduced by 20%. Find the sale price of a good

originally costing $580.

Solution

As a fraction

20% is the same as 20

100= 0.2

so the fall in price is

0.2 × 580 = 116

Hence the final price is

580 − 116 = $464

d) The cost of a refrigerator is £350.15 including sales tax at 17.5%.

What is the price of the refrigerator without sales tax?

Solution. To determine the price of the refrigerator without sales tax, we

divide £350.15 by the factor

1 +17.5

100= 1.175 .

So the price of the refrigerator without VAT is

350.15

1.175= £298.

Similarly, if a quantity decreases by a certain percentage, then that

percentage of the original quantity is subtracted from the original to

obtain its new value. The new value may be determined by multiplying

the original value by the quantity

1 −𝑟

100

e) A person’s income is K25,000 of which K20,000 is taxable. If the rate of

income tax is 22%, calculate the person’s net income.

Solution. The person’s net income comprises the part of his salary that is

not taxable (K5,000) together with the portion of his taxable income that

remains after the tax has been taken. The person’s net income is therefore


5,000 + (1 −22

100) × 20000 = 5,000 +

78

100× 20,000

= 5,000 + 78 × 200

= 5,000 + 15,600

= 𝐾20,600

Exercises

1. Calculate

(a) 10% of $2.90 (b) 75% of $1250 (c) 24% of $580

2. A calculator has been marked up 15% and is being sold for

$78.50. How much did the store pay the manufacturer of the

calculator?

3. A shirt is on sale for $15.00 and has been marked down 35%. How

much was the shirt being sold for before the sale?

4. A firm’s annual sales rise from 50 000 to 55 000 from one year to the

next. Express the rise as a percentage of the original.

5. The government imposes a 15% tax on the price of a good. How much

does the consumer pay for a good priced by a firm at $1360?

6. Investments fall during the course of a year by 7%. Find the value of

an investment at the end of the year if it was worth $9500 at the

beginning of the year.

7. If the annual rate of inflation is 4%, find the price of a good at the end

of a year if its price at the beginning of the year is $25.

8. The cost of a good is $799 including 17.5% VAT (value added tax).

What is the cost excluding VAT?

9. Express the rise from 950 to 1007 as a percentage.

10. Current monthly output from a factory is 25 000. In a recession, this is

expected to fall by 65%. Estimate the new level of output.

11. As a result of a modernization programme, a firm is able to reduce the

size of its workforce by 24%. If it now employs 570 workers, how

many people did it employ before restructuring?

12. Shares originally worth $10.50 fall in a stock market crash to $2.10.

Find the percentage decrease.


25

2.4 Powers and Indices

Let 𝑥 be a number and 𝑛 be a positive

integer, then 𝑥𝑛 denotes 𝑥 multiplied by

itself 𝑛 times. Here 𝑥 is known as the

base and 𝑛 is the power or index or

exponent. For example,

𝑥5 = 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥.

More specifically, if n is a positive integer, we have

1. 𝑥𝑛 = 𝑥 ⋅ 𝑥 ⋅ 𝑥 ⋅⋅⋅ 𝑥 3. 𝑥−𝑛 = 1

𝑥𝑛 = 1

𝑥 ⋅ 𝑥 ⋅ 𝑥 ⋅⋅⋅ 𝑥 for x≠ 0

2. 1

𝑥−𝑛 = 𝑥𝑛 for x≠ 0 4. 𝑥0 = 1

If 𝑟𝑛 = 𝑥, where n is a positive integer, then r is an nth root of x.

The Principal nth root of x is the nth root of x that is positive if x is

positive and negative if x is negative and n is odd.

We denote the principle nth root of x by √𝑥𝑛

For example

√92

= 3, √−83

= −2, and √1

27

3

=1

3

The symbol √𝑥𝑛

is called a radical.


There are rules for multiplying and dividing two algebraic expressions or

numerical values involving the same base raised to a power.

The table below gives the Laws of Exponents.

𝑳𝒂𝒘 𝑬𝒙𝒂𝒎𝒑𝒍𝒆

1 𝑥1 = 𝑥 61 = 6

2 𝑥0 = 1 70 = 1

3 𝑥−1 =

1

𝑥 4−1 =

1

4

4 𝑥𝑚𝑥𝑛 = 𝑥𝑚+𝑛 𝑥2𝑥3 = 𝑥2+3 = 𝑥5

5 𝑥𝑚

𝑥𝑛 = 𝑥𝑚−𝑛

𝑥6

𝑥2 = 𝑥6−2 = 𝑥4

6 (𝑥𝑚)𝑛 = 𝑥𝑚𝑛 (𝑥2)3 = 𝑥2×3 = 𝑥6

7 (𝑥𝑦)𝑛 = 𝑥𝑛𝑦𝑛 (𝑥𝑦)3 = 𝑥3𝑦3

8 (

𝑥

𝑦)

𝑛

=𝑥𝑛

𝑦𝑛 (

𝑥

𝑦)

2

=𝑥2

𝑦2

9 𝑥−𝑛 =

1

𝑥𝑛 𝑥−3 =

1

𝑥3

And the law about Fractional Exponents:


27

10

Example 2.1

Simplify the following using the rules of indices:

1. 𝑥2

𝑥3 2⁄ 2. 𝑥2𝑦3

𝑥4𝑦

Write down the values of the following without using a calculator

a) 3-3 b) 163

4 c) 16−3

4

d) 27−1

3 e) 43

2 f) 190

Rationalizing the denominator of a fraction is a procedure in which a

fraction having a radical in its denominator is expressed as an equal

fraction without a radical in its denominator.

Example 2.2

Rationalizing the Denominator

a. 2

√5 b.

2

√36 c.

√2

√3

Example 2.3

Exponents

a. Eliminate the negative exponents in 𝑥−2𝑦3

𝑧−2 for x ≠ 0, z ≠ 0.

b. Simplify 𝑥27

𝑥3𝑦5 for 𝑥 ≠ 0, 𝑧 ≠ 0.

c. Simplify (𝑥5𝑦8)5

d. Simplify (𝑥5

9⁄ 𝑦4

3⁄ )18

e. Simplify (𝑥1/5𝑦6/5

𝑧2/5 )5


Example 2.4

Radicals

a. Simplify √484

b. Rewrite √2 + 5𝑥 without using a radical sign.

c. Rationalize the denominator of √25

√63 and simplify.

d. Simplify √20

√5

2.5 Quadratic Equations

The simplest non-linear equation is known as a quadratic equation and

takes the form

𝑓 (𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

for some parameters a, b and c. In other words, a quadratic equation must

have a squared term as its highest power

For the moment we concentrate on the mathematics of quadratics and

show how to solve quadratic equations.

The values of x that satisfy the equation f(x) = 0 are known as the roots

or solutions of the equation. These two terms are used interchangeably.

Therefore, we say that 𝑥 = −2 and 𝑥 = 1/2 are the roots or solutions

of the quadratic equation 2𝑥2 + 3𝑥 − 2 = 0.

Consider the elementary equation

𝑥2 − 9 = 0

It is easy to see that the expression on the left-hand side is a special case

of the above ( 𝑓 (𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐) with a = 1, b = 0 and c = −9. To

solve this equation we add 9 to both sides to get

𝑥2 = 9

so we need to find a number, x, which when multiplied by itself produces

the value 9.

A moment’s thought should convince you that there are exactly two

numbers that work, namely

3 and −3 because

3 × 3 = 9 and (−3) × (−3) = 9

The name

Quadratic comes

from "quad"

meaning square,

because the

variable gets

squared (like x2).


29

These two solutions are called the square roots of 9. The symbol √ is

reserved for the positive square root, so in this notation the solutions are

√9 and −√9. These are usually combined and written±√9.

The equation

𝑥2 − 9 = 0

is trivial to solve because the number 9 has obvious square roots.

2.5.1 Solving Quadratic Equations

The solution of a quadratic equation is the value of x when you set the

equation equal to zero i.e. When you solve the following general

equation: 0 = 𝑎𝑥² + 𝑏𝑥 + 𝑐

There are a number of techniques for determining the roots of a quadratic

equation. Below are the four most commonly used methods to solve

quadratic equations.

The Quadratic Formula

Factoring

Completing the Square

Factor by Grouping

Solving by Factorization Method

If the expression defining a quadratic function can be factorized as a

product of linear factors, then equating each of the factors to zero and

solving the resulting linear equations will provide the roots.

Example 2.5.1

Solve 𝑥2 + 13𝑥 + 30 = 0 using factorization.

Solution.

First, we factorize the quadratic expression 𝑥2 + 13𝑥 + 30 as a product

of two linear factors (x + A) and (x + B), where A and B are two

constants that need to be determined. Since

(𝑥 + 𝐴)(𝑥 + 𝐵) = 𝑥2 + (𝐴 + 𝐵)𝑥 + 𝐴𝐵,

then the constants A and B need to be chosen so that

A + B = 13, AB= 30.

The possible combinations of integers whose product is 30 are 30 × 1, 15

× 2, 10×3, and 6×5. Of course, one also has the combinations in which

the integers have been negated such as (−30) × (−1), but out of these


combinations the only one for which the pair of integers sums to 13 is 10

× 3. Therefore, we choose

A = 10 and B = 3, i.e.,

𝑥2 + 13𝑥 + 10 = (𝑥 + 10)(𝑥 + 3).

We now solve the equation (x+3)(x + 10) = 0.

For the product of the two linear terms x+3 and x+10 to be zero, at least

one of them must be zero. So either

𝑥 + 3 = 0 𝑜𝑟 𝑥 + 10 = 0.

If x + 3 = 0 then x = −3, and if x + 10 = 0 then x = −10.

Therefore, the roots of the equation x2 + 13x + 30 = 0 are x = −3 and x =

−10.

Example 2

Solve the quadratic equation 2𝑥2 − 11𝑥 + 12 = 0 using factorization.

Solution.

As in the previous example, the first step is to factorize the quadratic

expression 2𝑥2 − 11𝑥 + 12 as a product of linear factors.

These linear factors must be of the form (2𝑥 + 𝐴) and (𝑥 + 𝐵) in order

to retrieve the quadratic factor 2 × 2, where A and B are two positive

constants.

Since

(2𝑥 + 𝐴)(𝑥 + 𝐵) = 2𝑥2 + (𝐴 + 2𝐵)𝑥 + 𝐴𝐵,

then the constants A and B need to be chosen so that

𝐴 + 2𝐵 = −11,

AB= 12.

The possible combinations of integers whose product is 12 are 12 × 1, 6 ×

2, 4 × 3, −4× −3, −6× −2, and −12 × −1.

The only pair of integers amongst these for which A + 2B = −11 is A =

−3 and B = −4. Therefore, we have

2𝑥2 − 11𝑥 + 12 = (2𝑥 − 3)(𝑥 − 4).

The problem now is to solve the equation

(2𝑥 − 3)(𝑥 − 4) = 0.

Either 2x − 3 = 0 or x − 4 = 0. If 2x − 3 = 0 then 2x = 3 and x = 3/2.

If x−4 = 0, then x = 4.

Therefore, the two roots of the equation 2𝑥2 − 11𝑥 + 12 = 0 are x =

3/2 and x = 4.


31

Most quadratic expressions, however, do not factorize easily in the sense

that they cannot be expressed as a product of linear factors with integer

coefficients, even if the coefficients of the quadratic equation are

integers.

For example, the quadratic equation 3𝑥2 − 9𝑥 + 5 = 0 cannot be

factored into a product of linear factors with integer coefficients. Clearly,

a more systematic approach is required.

The Quadratic Formula

A quadratic equation has the general form

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

where a, b and c are constants.

The solution of this equation may be found by using the formula,

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

The following examples describes how to use this formula. It also

illustrates the fact that a quadratic equation can have two solutions, one

solution or no solutions.

Example 2.5.2

Solve the quadratic equations

(a) 2𝑥2 + 9𝑥 + 5 = 0

(b) 𝑥2 − 4𝑥 + 4 = 0

(c) 3𝑥2 − 5𝑥 + 6 = 0

Solution

(a) For the equation

2𝑥2 + 9𝑥 + 5 = 0

we have a = 2, b = 9 and c = 5. Substituting these values into the formula

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Gives

𝑥 =−9 ± √92 − 4(2)(5)

2(2)

𝑥 =−9 ± √81 − 40

4


𝑥 =−9 ± √41

4

The two solutions are obtained by taking the + and − signs separately:

that is,

𝑥 =−9 + √41

4= −0.649

𝑥 =−9 − √41

4= −3.851

It is easy to check that these are solutions by substituting them into the

original equation. For example, putting x = −0.649 into

2𝑥2 + 9𝑥 + 5

gives

2(−0.649)2 + 9(−0.649) + 5 = 0.001 402

which is close to zero, as required. We cannot expect to produce an exact

value of zero because we rounded √41 to 3 decimal places. You might

like to check for yourself that −3.851 is also a solution.

(b) For the equation

𝑥2 − 4𝑥 + 4 = 0

we have a = 1, b = −4 and c = 4. Substituting these values into the

formula

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Gives

𝑥 =−(−4) ± √(−4)2 − 4(1)(4)

2(1)

=4 ± √16 − 16

2

=4 ± √0

2

=4 ± √0

2

Clearly we get the same answer irrespective of whether we take the + or

the − sign here. In other words, this equation has only one solution, x = 2.

As a check, substitution of x = 2 into the original equation gives

(2)2 − 4(2) + 4 = 0

c) For the equation

3𝑥2 − 5𝑥 + 6 = 0

we have a = 3, b = −5 and c = 6. Substituting these values into the

quadratic formula gives


33

𝑥 =−(−5) ± √(−5)2 − 4(3)(6)

2(3)

=5 ± √25 − 72

6

=5 ± √(−47)

6

The number under the square root sign is negative and it is impossible to

find the square root of a negative number. We conclude that the quadratic

equation

3𝑥2 − 5𝑥 + 6 = 0

has no real solutions.

Practice Problem

Solve the following quadratic equations (where possible):

(a) 2𝑥2 − 19𝑥 − 10 = 0 (b) 4𝑥2 + 12𝑥 + 9 = 0

(c) 𝑥2 + 𝑥 + 1 = 0 (d) 𝑥2 − 3𝑥 + 10 = 2𝑥 + 4

Example 2.5.3

Solve the quadratic equation 3𝑥2 − 9𝑥 + 5 = 0 using the formula.

Solution.

First we compare the coefficients of this equation with those of the

general quadratic equation.

If we do this, we notice that 𝑎 = 3, 𝑏 = −9, and 𝑐 = 5.

Inserting these values into the quadratic formula gives

𝑥 =−(−9) ± √(−9)2 − 4 × 3 × 5

2 × 3

=9 ± √81 − 60

6

=9 ± √21

6

Note that 21 is not a perfect square, and therefore the roots of this

equation can only be expressed in decimal representation to a specified

number of decimal places. Therefore, to four decimal places the two

solutions of this equation are

x =9+√21

6= 2.2638 and, x =

9−√21

6= 0.7362


Completing the Square Method

Some quadratics cannot be easily factorized so we have to use the

technique of "Completing the Square."

The Idea behind completing the square is to rearrange the quadratic into

the neat "(squared part) equals (a number)" format. i.e (𝑥 − 4)2 = 23.

We show this method using an example.

This is the original problem. 4x2 – 2x – 5 = 0

Move the loose number over to the other

side. 4x2 – 2x = 5

Divide through by whatever is multiplied

on the squared term.

Take half of the coefficient (don't forget

the sign!) of the x-term, and square it. Add

this square to both sides of the equation.

Convert the left-hand side to squared

form, and simplify the right-hand side.

(This is where you use that sign that you

kept track of earlier. You plug it into the

middle of the parenthetical part.)

Square-root both sides, remembering the

"±" on the right-hand side. Simplify as

necessary.

Solve for "x =".

Remember that the "±" means that you

have two values for x.

Solve x2 + 6x – 7 = 0 by completing the square.

Do the same procedure as above, in exactly the same order.

This is the original equation. x2 + 6x – 7 = 0


35

Move the loose number over to the other side. x2 + 6x = 7

Take half of the x-term (that is, divide it by two) (and

don't forget the sign!), and square it. Add this square to

both sides of the equation.

x2 + 6x +9 = 7+9

Convert the left-hand side to squared form. Simplify

the right-hand side. (x + 3)2 = 16

Square-root both sides. Remember to do "±" on the

right-hand side. x + 3 = ± 4

Solve for "x =". Remember that the "±" gives you two

solutions. Simplify as necessary.

x = – 3 ± 4

= – 3 – 4, –3 + 4

= –7, +1

Solutions of a Quadratic Equation

The number of solutions of a quadratic equation depends on the sign of

the expression under the square root sign in this formula. A quadratic

equation has two, one or no solutions depending on whether the

expression 𝑏2 − 4𝑎𝑐 is positive, zero, or negative:

If 𝑏2 − 4𝑎𝑐 > 0, there are two solutions

𝑥 =−𝑏+√𝑏2−4𝑎𝑐

2𝑎 and 𝑥 =

−𝑏−√𝑏2−4𝑎𝑐

2𝑎.

If 𝑏2 − 4𝑎𝑐 = 0, then there is one solution

𝑥 = −𝑏

2𝑎

If 𝑏2 − 4𝑎𝑐 < 0, then there are no solutions since the square root of

𝑏2 − 4𝑎𝑐 does not exist in this case.

2.6 Absolute Value

On a number line, the distance of a number x from 0 is called the absolute

value of x and is denoted by |𝑥|. For example, |−12| = |12| =12 because both12 and -12 are 12 units from 0.

Similarly,|0| = 0. Notice that |𝑥| can never be negative.


For any real number x the absolute value or modulus of x is defined as

As can be seen from the above definition, the absolute value of x is

always either positive or zero, but never negative.

Since the square-root notation without sign represents the positive square

root, it follows that

Solving Absolute Value Equations

The absolute value property (that both the positive and the negative

become positive) makes solving absolute-value equations a little tricky.

But once you learn the "trick", they're not so bad. Let's start with

something simple:

Solve | x | = 3

From the definition of a modulus, | 3 | = 3 and | –3 | = 3, so x must be 3 or

–3.

Then the solution is x = –3, 3.

Solve | x + 2 | = 7

To clear the absolute-value bars, we must split the equation into its two

possible two cases, one case for each sign:

(x + 2) = 7 or –(x + 2) = 7

x + 2 = 7 or –x – 2 = 7

x = 5 or –9 = x

Then the solution is x = –9, 5.

Solve | x2 – 4x – 5 | = 7

First, I'll clear the absolute-value bars by splitting the equation into its

two cases:

( x2 – 4x – 5 ) = 7 or –(x2 – 4x – 5) = 7

Solving the first case, we get:

x2 – 4x – 5 = 7

x2 – 4x – 12 = 0

(x – 6)(x + 2) = 0

x = 6, x = –2

Solving the second case, we get:

Absolute-value

equations

always work this

way: to be able

to remove the

absolute-value

bars, you have

to isolate the

absolute value

onto one side,

and then split

the equation

into the two

possible cases.


37

–x2 + 4x + 5 = 7

–x2 + 4x – 2 = 0

0 = x2 – 4x + 2

Applying the Quadratic Formula to the above, we get:

Then my solution is:

𝑥 = −2, 6, 2 ± √2

Practice Problems

1. Solve a) |2𝑥 + 1| = 3

b) |𝑥| < |𝑥 + 4|

c) |13 − 2𝑥| ≤ 5.

2. In the manufacture of widgets, the average dimension of a part is 0.01

cm. Using the absolute-value symbol, express the fact that an

individual measurement x of a part does not differ from the average by

more than 0.005 cm.

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