BBA120 Business Mathematics

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


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.

3

4 Unit 2: Equations and Inequalities

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

Remember

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.

It is useful to remember that a minus sign is a -1, so -5 is read as −1 × 5

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.

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

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.


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)

= 𝑥

2

+ 4𝑥 − 3𝑥 − 12

= 𝑥

2

+ 𝑥 − 12

2.3 Order of Operations

Note: Other texts use the Acronyms

BODMAS or

PEMDAS.

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

5


Exponent next: square the 2 or 2

2

= 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

Note that in the fraction

𝑎 𝑏

a can be greater than b

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,





5

1

3

of students in a lecture may be female or

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.

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.


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

7


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 𝑥

2

−5𝑥 e)

(

(

2

3

5

)

11

)

= (

2

3

) (

11

5

) =

22

15

f)

2𝑥 𝑥+𝑦

3𝑥

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:

In some text books, a bar notation is used instead of a dot e.g.

52.71656565…

(65 repeating infinitely often) may be written

52.7165


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

9


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


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:

11


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)


= 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

The solution of an equation is simply the value or values of the unknown(s) for which the lefthand side (LHS) of the equation is equal to the righthand side (RHS).

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.

13


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

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

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

1.5 = 𝑥

(dividing both sides by 6)

(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


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).

15


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

-10

0

10

22

(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


(c)

2𝑥 – 6 < 12 – 4𝑥

2𝑥 + 4𝑥 − 6 < 12

6𝑥 < 12 + 6

6𝑥 < 18 𝑥 < 3 add 4x to both sides add 6 to both sides dividing both sides by 6



Solve x

2

– 3x + 2 > 0

First, we have to find the x-intercepts of the associated quadratic equation, because the intercepts are where 𝑦 = 𝑥

2

– 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 𝑥

2

– 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:

17

18 Unit 2: Equations and Inequalities 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 = 𝑥

2

– 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 = x

2

– 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.


Then the solution of x

2

– 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.

19


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)

2𝑥

2

− 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.


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,

21


25

25% of 140 is

100

× 140 = 35 and

4

4% of 5, 200 is

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


12

12% is the same as

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

20% is the same as

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

23


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.


2.4 Powers and Indices

Let 𝑥 be a number and 𝑛 be a positive integer, then 𝑥 𝑛

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,

𝑥

5

= 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥.

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

1. 𝑥 𝑛

= 𝑥 ⋅ 𝑥 ⋅ 𝑥 ⋅⋅⋅ 𝑥 3. 𝑥

−𝑛

=

1 𝑥 𝑛

=

1 𝑥 ⋅ 𝑥 ⋅ 𝑥 ⋅⋅⋅ 𝑥

for x≠ 0

2.

1 𝑥

−𝑛

= 𝑥 𝑛

for x≠ 0 4. 𝑥

0

= 1

If 𝑟 𝑛

= 𝑥, where n is a positive integer, then r is an n

th

root of x.

The Principal n

th

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

√𝑥

For example

2

√9

= −2, and √

1

27

=

1

3

The symbol 𝑛

√𝑥

is called a radical.

25


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 𝑥

1

= 𝑥

2 𝑥

0

= 1

61 = 6

70 = 1

3 𝑥

−1

=

1 𝑥

4

−1

=

1

4

4 𝑥 𝑚 𝑥 𝑛

= 𝑥 𝑚+𝑛 𝑥

2 𝑥

3

= 𝑥

2+3

= 𝑥

5

5 𝑥 𝑚 𝑥 𝑛

= 𝑥 𝑚−𝑛

6

(𝑥 𝑚

) 𝑛

= 𝑥 𝑚𝑛 𝑥

(𝑥

2

6 𝑥 2

= 𝑥

6−2

)

3

= 𝑥

= 𝑥

2×3

4

= 𝑥

6

7 (𝑥𝑦) 𝑛

= 𝑥 𝑛 𝑦 𝑛

8

( 𝑥 𝑦

) 𝑛

= 𝑥 𝑛 𝑦 𝑛

(𝑥𝑦)

3

= 𝑥

3 𝑦

( 𝑥 𝑦

)

2

= 𝑥

2 𝑦 2

9 𝑥

−𝑛

=

1 𝑥 𝑛 𝑥

−3

=

1 𝑥 3

And the law about Fractional Exponents:

3


10

Example 2.1

Simplify the following using the rules of indices:

1. 𝑥 𝑥

2

2

. 𝑥

2 𝑥

4 𝑦 𝑦

3

Write down the values of the following without using a calculator a)

3

-3 b)

16

3

4 c) 16

−

3

4 d)

27

−

1

3 e)

4

3

2 f)

19

0

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

2

b.

6

√3

c.

√2

√3

Example 2.3

Exponents a. Eliminate the negative exponents in

𝑥

−2

𝑦

3

𝑧

−2 for x ≠ 0, z ≠ 0.

𝑥

2

7

b. Simplify

𝑥

3

𝑦

5 for 𝑥 ≠ 0, 𝑧 ≠ 0. c. Simplify ( 𝑥

5 𝑦

8 ) 5 d. Simplify

(𝑥

5

⁄

9 𝑦

4

⁄

3

)

18 e. Simplify

(

𝑥

1/5

𝑦

6/5

𝑧

2/5

)

5

27


Example 2.4

Radicals a. Simplify

4

√48 b. Rewrite

√2 + 5𝑥 without using a radical sign. c. Rationalize the denominator of

5

√2

3

√6

and simplify. d. Simplify

√20

√5

2.5 Quadratic Equations

The name

Quadratic comes from "quad" meaning square, because the variable gets squared (like x

2

).

The simplest non-linear equation is known as a quadratic equation and takes the form 𝑓 (𝑥) = 𝑎𝑥

2

+ 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

2𝑥

2

+ 3𝑥 − 2 = 0.

Consider the elementary equation 𝑥

2

− 9 = 0

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

𝑓 (𝑥) = 𝑎𝑥

2

+ 𝑏𝑥 + 𝑐) with a = 1, b = 0 and c = −9. To solve this equation we add 9 to both sides to get 𝑥

2

= 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


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 𝑥

2

− 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 𝑥

2

+ 13𝑥 + 30 = 0 using factorization.

Solution.

First, we factorize the quadratic expression 𝑥

2

+ 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) = 𝑥

2

+ (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

29


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., 𝑥

2

+ 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

2𝑥

2

− 11𝑥 + 12 = 0 using factorization.

Solution.

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

2𝑥

2

− 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) = 2𝑥

2

+ (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

2𝑥

2

− 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

2𝑥

2

3/2 and x = 4.

− 11𝑥 + 12 = 0 are x =


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

3𝑥

2

− 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 𝑎𝑥

2

+ 𝑏𝑥 + 𝑐 = 0 where a, b and c are constants.

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

−𝑏 ± √𝑏 2 − 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)

2𝑥

2

+ 9𝑥 + 5 = 0

(b)

𝑥

2

− 4𝑥 + 4 = 0

(c)

3𝑥

2

− 5𝑥 + 6 = 0

Solution

(a) For the equation we have a = 2, b = 9 and c = 5. Substituting these values into the formula 𝑥 =

−𝑏 ± √𝑏 2 − 4𝑎𝑐

2𝑎

Gives

2𝑥

2

+ 9𝑥 + 5 = 0 𝑥 =

−9 ± √9

2

− 4(2)(5)

2(2) 𝑥 =

−9 ± √81 − 40

4

31


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

2𝑥

2

+ 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 𝑥

2

− 4𝑥 + 4 = 0 we have a = 1, b = −4 and c = 4. Substituting these values into the formula 𝑥 =

−𝑏 ± √𝑏

2

2𝑎

− 4𝑎𝑐

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

3𝑥

2

− 5𝑥 + 6 = 0 we have a = 3, b = −5 and c = 6. Substituting these values into the quadratic formula gives


𝑥 =

−(−5) ± √(−5)

2

2(3)

− 4(3)(6)

=

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

3𝑥

2

− 5𝑥 + 6 = 0 has no real solutions.

Practice Problem

Solve the following quadratic equations (where possible):

(a)

2𝑥

2

− 19𝑥 − 10 = 0

(b)

4𝑥

2

+ 12𝑥 + 9 = 0

(c)

𝑥

2

+ 𝑥 + 1 = 0

(d)

𝑥

2

− 3𝑥 + 10 = 2𝑥 + 4

Example 2.5.3

Solve the quadratic equation

3𝑥

2

− 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

33


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. 4x

2

– 2x – 5 = 0

Move the loose number over to the other side.

4x

2

– 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 x

2

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

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

This is the original equation.

x

2

+ 6x – 7 = 0


Move the loose number over to the other side.

x

2

+ 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.

x

2

+ 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 𝑏

2

− 4𝑎𝑐 is positive, zero, or negative:



If 𝑏

2

− 4𝑎𝑐 > 0, there are two solutions 𝑥 =

−𝑏+√𝑏

2

−4𝑎𝑐

2𝑎 and 𝑥 =

−𝑏−√𝑏

2

2𝑎



If 𝑏

2

− 4𝑎𝑐 = 0, then there is one solution

−4𝑎𝑐

. 𝑥 = − 𝑏

2𝑎



If 𝑏

2 𝑏

2

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

− 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.

35


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

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.

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 | x

2

– 4x – 5 | = 7

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

( x

2

– 4x – 5 ) = 7 or –(x

2

– 4x – 5) = 7

Solving the first case, we get:

x

2

– 4x – 5 = 7

x

2

– 4x – 12 = 0

(x – 6)(x + 2) = 0

x = 6, x = –2

Solving the second case, we get:


–x

2

+ 4x + 5 = 7

–x

2

+ 4x – 2 = 0

0 = x

2

– 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.

37

BBA120 Business Mathematics

SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT

BBA120 BUSINESS MATHEMATICS / MATHEMATICAL ANALYSIS

Contents

Unit 2: Equations and Inequalities 3

Unit 2: Equations and Inequalities

2.0 Arithmetic Operations

2.1. Addition and subtraction

Examples

2.2 Multiplication and Division

Examples

2.3 Order of Operations

Note: Other texts use the Acronyms

BODMAS or

PEMDAS.

Examples

b) 23 × 3 + (5 − 1).

c) 4 – 3[4 – 2(6 – 3)] ÷ 2.

2.4 Fractions

Note that in the fraction

a can be greater than b

2.4.1 Addition and Subtraction of Fractions

Example

2.4.2 Multiplication and Division of fractions

Examples

(

(

)

)

2

3

11

5

22

15

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

In some text books, a bar notation is used instead of a dot e.g.

52.71656565…

(65 repeating infinitely often) may be written

52.7165

2.5.1 Converting Decimals to Fractions

2.5.2 Converting repeating decimals to fractions

Example #1:

Example #2:

2.5.2 Significant Figures

2.1 Linear Equations

The solution of an equation is simply the value or values of the unknown(s) for which the lefthand side (LHS) of the equation is equal to the righthand side (RHS).

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

2.2 Simple Inequalities

2.2.1 Solving Inequalities

Rational Inequalities

Exercise

2.3 Percentages

2.4 Powers and Indices

2

√5

2

√3

√2

√3

𝑥

𝑦

𝑧

𝑥

7

𝑥

𝑦

𝑥

𝑦

𝑧

5

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