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Student Name: Aseel Mohammed AlSawaii

ID Number: 1020137

Introduction

Algebraic equations allow people to solve for quantities that are not known. Many times it is not

possible to directly figure out how much something is through a measurement alone. You might

also want to predict what something is without building everything necessary to find out. If you

know, for example that something costs $10, and you want to know how many you could buy

with $500, you can set up an equation, 10X = 500. Ten represents the rate, and X represents the

unknown. To solve any equation, mathematicians use various methods to use known information

to find unknown information. They set up the information in a number sentence, and then do

inverse operations to isolate the unknown. This essay will talk about how to solve the three most

common types of single variable equations: linear equations, quadratic equations, and

exponential equations.

All equations are built out of quantities and variables. Variables are unknown so they cannot be

represented by exact numbers. Unknowns are usually represented by the letter X but it helps the

person solving the equation if they use letters that hint at the answers. For example, if you were

trying to solve for how many Widgets are going to be needed, the variable could be written as W.

They are called equations because one quantity, also known as an expression, is equivalent to

another expression. If for example we say that six cereal boxes cost forty-two dollars, we could

write this as an equation: 6C = 42. The unknown, C, denotes the price of the Cereal box. It is

obvious that six times whatever the price is is equivalent to the total 42. So both expressions are

placed at opposite sides of the equal sign. Whenever steps are taken to solve any equation,

whatever is done to one side of the equation must be done to the other side in exactly the same

way. If this does not happen, the equivalent relationship between the two expressions will be

changed. For example, if you added 5 to one side and not the other, you would get 6C = 47. The

expression would then say they price of six cereal Boxes is forty-seven dollars instead of fortytwo. This would be a different problem. If you added 5 to both sides of the equation, you would

get 6C + 5 = 47, which is differently phrased but still expresses the same relationship as before.

Five more than the 6 boxes of cereal is five more than forty-two, or forty seven.

For this reason, when one is to solve an equation, one must do the same operations to both sides

of the equation in order to isolate the variable. If this is done, the equivalence and the expression

is maintained. The complexity is simplified. Let us solve the equation already discussed as an

example: 6C + 5 = 47. The C value on the right needs to be isolated so that the value can be

determined. There is a 6 and a 5 on that side that must be removed. It would change the

relationship if one just crossed them out. The first thing one should do is subtract the five on both

sides. This helps isolate the variable. 6C + 5 – 5 = 47 – 5. Numbers go with numbers and

variable expressions go with variable expressions that are like them. This is called combining

like terms and must be done so that equations become easy to solve. These becomes 6C = 42 and

the equation is simplified. Notice what happened here: The plus 5 and the minus 5 “Cancelled”

each other out, or became eliminated as an obstacle to solving on one side. People who solve

equations should be trying to cancel all quantities out that are not the variable so that the value of

the variable can be determined. Also notice that since we were adding 5, to remove it, we would

have to subtract 5. This is called doing the “inverse operation. If the quantity you want to remove

is being added, you will want to subtract it. If it is being subtracted, you will want to add it. If a

quantity is dividing the variable, one must multiply to remove it. If it is being multiplied, as the 6

is in 6C, you will want to divide by 6 on both sides to cancel out that value. Six divided by six

gives one, and one times any quantity is itself. If in the 6C = 42 example, we divide both sides by

6, you end up with C = 7. Seven dollars is the price. One can check if this is right by replacing

the C with the value one solved for, the seven. 6C + 5 = 47 becomes Six times Seven plus five =

47. Do the math and you find 47 is the same as 47. The equation is equal on both sides, and

therefore the equivalence relationship has been maintained.

It does not matter which operation someone does on an equation so long as every number in it is

affected properly. Practice allows for the student to solve questions in the fewest steps and not

overcomplicate things. In the first example, 6C + 5 = 47, one could divide by 6 first and get C +

5/6 = 47/6. This is still correct; if one subtracted 5/6ths on both sides now, one would arrive at C =

42/6 which is 7, the correct answer. But it might lead to confusion since the problem has been

made more difficult.

People solving equations should take care to follow the standard order of operations or they will

solve everything incorrectly. Everything written in a parenthesis should be done first. Any

exponent should be taken into account next. Next is multiplication and division, which can be

done in any order along the equation so long as care is taken not to forget to multiply or divide a

certain term. Addition and Subtraction come last. If one needs some part of the equation to

happen before other parts, one would be wise to insert that section into brackets in order to make

sure the order is correct. For example in this equation 3(4+X) = 48 (3-2), one has to combine

each expression in a pair of parentheses and treat it as one number. Three minus two is one so

that parenthesis is taken care of. One times forty-eight is itself. The X + 4 term cannot be

simplified since no one knows what the X is. In this problem, one has to multiply everything in

the parenthesis by three. The result is 12 + 4X = 48. Everything can be solved with inverse

operations. Subtractive twelve on both sides creates 4X = 36. Then, after dividing by four, X = 9

is revealed.

Graphs of equations are pictures of all the answers X could be if it did not equal an exact value.

Linear equations, single degree equations, form a straight line when graphed. This happens

because they increase or decrease at a constant rate. The unknown is multiplied by a constant

number called the slope. A negative slope means the value falls as the unknown variable gets

bigger. A positive slope means the value rises as the unknown variable increases. A negative

slope means it decreases. The graph of a quadratic equation forms a shape called a parabola. This

U shape goes down then up or up then down in a U shape that mimics the arc of a thrown ball.

The U shape happens because, unlike linear equations, the unknown is multiplied by itself for at

least one term, to produce X squared, a second degree equation. Ever-increasing numbers

multiplied by other increasing numbers grow at much faster rates than straight lines do and they

produce a curve. Exponential equations, where a base number is multiplied by itself an unknown

number of times, increase at even quicker rates. Their curve rises more steeply with each

increment. This is called an exponential rise. Quadratics and exponentials need to be solved

slightly different since they are not linear and do not increase by simple means of dividing by the

slope.

Quadratic equations are called these because they can be solved in a box with four quadrants.

Because, being U shaped, they can hit any horizontal line they cross twice, there are two answers

to a quadratic equation. People usually set these Quadratic equations to zero since reduces the

number of factors a term like X² can be generated from and makes the problem easier. An

infinite number of numbers, decimals, and fractions can be multiplied together to get a number

like four. However, only multiplying by zero gets you to zero. Thus, every quadratic equation

should be rearranged to equal zero before it is solved. It turns into what is known as AX² + BX +

C = 0 which is the standard form. A, B, and C, are coefficients, or constant values the variables

are multiplied by. X is the variable that needs to be solved for.

There are three methods for solving a quadratic equation. The first is completing the square; the

others are based off of completing the square. It is a well-known fact that a square has all equal

sides, and that if you multiply the length of one side by itself you get the area of the square. It is

also known that there are two square roots of any number—a positive and a negative version—

and that the area of a square is equal to all of its parts if the square is cut up. All of these facts are

used. The quadratic equation—let us use 2X² + 26X + 84 = 0 as an example—has to be

decomposed into some quantity (X + N)² = M, so that if you take the square root of both sides,

you get (X + N) = ± The Square root of M. It is much easier to take the square root of (X+N)²

than the original quadratic equation since there is no method for the latter, whereas the former is

trivial. After this, the N value can easily be subtracted from both sides and the unknown is

revealed as X = – N ± the square root of M. Two values of X are produced.

Completing the square should be visualized. One can draw a square divided in four like a

window pane or checkerboard. Each side of the whole square is X + N long. The Upper left

square would be X long and X high. The Upper right square would be N long and X high. The

Lower Left square would be N high and X long. The Lower Right square is N long and N high.

The areas of each square are the various lengths and heights multiplied together. When you

multiply (X+N) by itself, you get four terms: X², NX, NX, and N². These four add up to the total

area of the square, which is M. Often when we have an equation, we start with only three of

these values: X², NX, NX. In order to complete the square, as per the name, one needs to solve

for the N² term. This is done through some trickery. In a general case (X + A) square is the same

as X² + 2AX + A². If in our quadratic equation all that is there is the X² plus 2AX term, we can

add the necessary A² term on both sides in order to make it work. A shortcut is to halve the B

term of the equation, then square the result and add that on the other side.

This is made clearer with an example. In 2X² + 26X + 84 = 0, the equation can be made simpler

if one divides everything by two to get X² + 13X + 42 = 0. The equation cannot be decomposed

if there’s a number in front of the X² term. Completing the square also does not work if the area

of the square is equal to zero or is negative. You need to subtract the numerical term on both

sides so that only the terms with variables are on one side. This produces X² + 13X = -42. The

X² + 13X has to be converted into an (X + N)² version now. As mentioned before, (X + N)², what

we want, is equivalent to X² + 2NX + N². We would have to make the 2N equal to 13 just so that

it can have a chance of matching with the X² + 13X. But as 13 is double N and not N, we would

have to say that N is half of thirteen (13/2 or 6.5). Notice too that (X + 13/2)² is not X² + 13X but

X² + 13X + 169/4. 13/2 times itself is 169/4. We cannot replace the original X² + 13X with the

simpler (X + 13/2)² without adding the 169/4 on both sides. Adding this value is “completing the

square.” One gets (X + 13/2)² = -42 + 169/4 which is ¼. Taking the square root of (X + 13/2)² =

¼ on both sides reveals X + 13/2 = ±1/2. Subtract the 13/2 on both sides to reveal X = -13/2 ± ½.

This creates two answers: X = -6 and -7 (X = -13/2 + ½ = -6 and X = -13/2 – ½ = -7).

If one takes the AX² + BX + C = 0 form and completes the square with it, it produces the

Quadratic Formula which can be used to solve any quadratic equation. The formula is X = [-B ±

√(B² – 4AC)] / 2A. This formula is best memorized. To use it, you change any quadratic into the

standard form, such as 2X² + 26X + 84 = 0 and then replace the relevant coefficients into the

formula. In this example, it would produce X = [ -26 ± √(26² – 4(2)(84))] / 2(2). The calculation

reveals two answers: X = -6 and -7.

The third way is factoring, which is to parse the quadratic equation into two quantities multiplied

together that make a sort of rectangle when multiplied instead of a square like before. There

might be different factors that multiply together to make zero. For this to work, the graph of the

quadratic equation has to touch or cross the X Axis, the same line that represents the zero the

equation is equal to. It also needs to hit the X axis at a rational number for it to split easily. If the

graph never crosses zero, the function is not factorable. Factoring is actually like multiplying

binomial terms like (X + A) backwards. We know (X + A) times (X + B) equals X² + AX + BX

+ AB. One already knows the numerical term in X² + 13X + 42 = 0: 42. One needs to factor the

+42 into negative and positive factors, and then choose which pair adds up to the B term. In this

example, the factors of 42 are either 1 & 42, -1 & – 42, -6 & -7, or 6 & 7. Six and seven add up to

thirteen, so you can insert them into the (X + A) times (X + B), all of them set equal to zero. One

would get (X+6) times (X+7). As the multiplied factors equal zero, one of these factors has to be

Zero, otherwise the quadratic equation would not make sense. So we set both factors equal to

zero and solve for them. If X + 6 = 0, X = -6 and as X + 7 = 0, X = -7. These factors can be

broken up only when it zero, since nothing else works so easily.

Exponential equations are the hardest to solve since the equations involves a constant being

raised to an unknown quantity, such as 2^X or 3^X. The inverse operation to an exponent is a

logarithm. The logarithm of, say 4, means the exponent necessary to turn a ten into a four

(0.60206). Calculators or logarithm tables tell you what these are. They can be deduced but the

process is very difficult and time consuming. The rules of logarithms include a rule called a

power rule where if you take a logarithm of 2^3, it becomes three times the logarithm of the

base, two. This makes sense because three times the exponent required to turn a ten into a two

would be the same as the exponent required to turn a ten into an eight, the value of three cubed.

If one has an equation of 2^x + 6 = 20, one has to combine like terms to make the problem

simple and then use logarithms to solve for the variables. There are no simple rules to simplify

something like the logarithm of (2^x + 6), only 2^x. One should in this example subtract six on

both sides to get 2^x = 14 and then take the logarithm of both sides. The power rule transforms

the left side into X times the Logarithm of 2, and the right side becomes the Logarithm of 14.

The Logarithms of two and fourteen are complex decimals but they are just numbers, so you can

divide by the Logarithm of 2 on both sides to determine that X is the quotient of the Logarithm

of 14 and the Logarithm of 2. Sometimes answers like this can be simplified using other rules of

logarithms. The logarithm of 14 can be rewritten as the Logarithm of 7 plus the Logarithm of 2.

When this sum is divided by the Logarithm of 2, the result is one plus Log 7 divided by Log 2.

To proceed further, one must use a calculator or a logarithm table to calculate the exact value of

3.807354. As with all equations, one can reinsert the answer into the original equations to check

for equivalence. 2^3.807354 is 14 and fourteen plus six is 20.

A much harder example is solving equations with different bases, such as A) 3^(X+2) = 4^(2-x),

or B) 2^x +5^x = 40. Equation A can be solved by taking the logarithms on both sides and using

the power rule again. One gets (X+2) x Log 3 = (2-X) x Log 2. Some division combines like

terms to arrive at (X+2) / (2-X) = Log 4 / Log 3. One has to simplify the right side as a number,

1.26186, then multiply both sides by (2-x) to remove the fraction. This gets X + 2 = 2.5237 –

1.26186 X, which can be solved like a normal linear equation, arriving at X = 0.2315. This is

also equivalent to -2 Log (3/4) / Log 12. Equation B can be solved by graphing the function and

interpolating when 2^x + 5^x is the same as 40. It is at around 2.215. There is no simple way to

combine and simplify 2^x and 5^x with logarithms.

Conclusion

This essay has explained many techniques for solving equations from linear equations and

quadratics to exponentials. The basic method of isolating the variable in question is a constant

throughout. Quadratic Equations and Exponentials involve a lot of rearranging and a few

formulae. But it is important to remember that math at its heart is just arithmetic with trends that

people can discern. Being able to do such math allows people to predict how things that do not

exist and have not happened yet will happen. These are only the basics of math, but they are vital

if the reader intends to plan things or predict what might happen before it happens.

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FACULTY OF BUSINESS AND MANAGEMENT STUDIES

UNDERGRADUATE BUSINESS PROGRAMME

ACADEMIC YEAR 2019-2020, 2nd Semester

August – 2020

Writ 1

Programme Title: A&F, BE

Module Title: Quantitative Methods

Module Code: GEC4003

Assessment Method: Assignment

Level: 3

Block: 1

Module Credits: 20

Weighting: 50%

Due Date: 23-Jun-2020, 12:00 PM Word Count: 3,000

Examiner(s): Dr. Romeo & Mr. Ahmad

Version: 1

Gulf College – Faculty of Business and Management Studies – In academic

Affiliation with CARDIFF SCHOOL OF MANAGEMENT

A.

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I affirm that this has been researched and completed in accordance with the college rules and

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I acknowledge that I read and understand the plagiarism guide written at the end of this

assessment. Any academic misconduct will be handled according to the rules and regulations of the

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Assessment Details

This Assignment comprises 50% of the total assessments marks. This will develop the following

skills:

Analytical skills: The assignment enables the students to analyse and evaluate the benefits

of the different activities done in the class.

Research skills. This skill will eventually help them understand and evaluate the meaning

and the basic concept of linear, quadratic, exponential and logarithmic equations.

In addition, the Assignment will test the following learning outcomes:

Apply basic mathematical methods

Demonstrate understanding of the relationship between equations and linear and nonlinear diagrams

Assignment Task

Using references to relevant theories, concepts and appropriately chosen examples about

different types of equations, write an assignment of 3000 words for the following topic.

1. The concept and nature of simple algebraic, quadratic and exponential equations.

2. The solution to simple algebraic equation. Discuss how to apply solution in solving a simple

algebraic equation.

3. The solution to quadratic equation using the three basic methods: factoring, quadratic, the

quadratic formula, and completing the square. Discuss how to apply all of the methods in

solving a quadratic equation.

4. The solution to exponential equation. Discuss how to apply solution in solving exponential

equation.

Your report should contain the following:

Introduction: Overall plan of the work with a reference to the discussion about the meaning

and concept of simple algebraic, quadratic, exponential equations. (400 words equivalent)

Solution and Discussion

Solution to simple algebraic equations as well as the manner how to check or verify to

determine if the answer is correct. Student needs to provide two (2) examples, one with

brackets and the other one without bracket and discuss the solution and the manner of checking the

answers. (500 words equivalent)

Solution to the three basic methods in solving quadratic equation as well as the manner

how to check or verify to determine if the answer is correct . Student needs to provide one

example for each with appropriate discussion of the solution and the manner of checking

the answers. (1,200 words equivalent)

Solution to exponential equations as well as the manner how to check or verify to

determine if the answer is correct. Student needs to provide two examples: one with

common base and one with different bases including appropriate discussion of the solution

and the manner of checking the answers. (500 words equivalent)

Conclusion – Student needs to summarise the key points in the discussion. (400 words

equivalent)

***END OF ASSIGNMENT TASK***

Guidelines to Students:

Diagrams are an effective and efficient method of conveying information. To obtain good

marks it will be necessary to include diagrams where relevant to illustrate your discussion.

Make sure that you incorporate these diagrams fully into your reports. In other words make

sure you refer to them in your text. Information should be drawn from respected reputable

academic sources such as textbooks and academic journals, and reports from economic

think tanks.

GGEC4003 – Quantitative Methods

AY: 2019-2020 / 2nd Semester

Marking Scheme

Question

No.

Description

Marks

Allocated

Introduction – Meaning and concept of :

1

Simple algebraic equation (5 marks)

Quadratic equation (5 marks)

Exponential equation (5 marks)

15

Solution to simple algebraic equation

2

Without brackets – Example with solution and checking (5 marks)

Discussion (5 marks)

20

With brackets – Example with solution and checking (5 marks)

Discussion (5 marks)

Solution of the three basic methods in solving quadratic equation

Factoring – Example with solution and checking (6 marks)

Discussion (4 marks)

3

Quadratic formula-Example with solution and checking (6 marks)

Discussion (4 marks)

30

Completing the square – Example with solution and checking (6 marks)

Discussion (4 marks)

Solution to exponential equation

4

With common base – Example with solution and checking (6 marks)

Discussion (4 marks)

20

Without common bases – Example with solution and checking (6 marks)

Discussion (4marks)

Conclusion – Summary of the significant points

5

Simple algebraic equation (4 marks)

Quadratic equation (7 marks)

Exponential equation (4 marks)

Total

15

100

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