Business and Management Studies Discussion

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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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Written work
A signed declaration that the work is your own (apart from otherwise referenced
acknowledgements) must be included after the reference page of your assignment
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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 the advice given by the module tutors on proper referencing to avoid plagiarism and
the rules on the academic unfair practice.
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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[Name of Student]
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Assessment Details
This Assignment comprises 50% of the total assessments marks. This will develop the following
skills:
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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:
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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.
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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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