SMK SULTAN ISMAIL, JOHOR BAHRU.

Name : Ainnatul Athirah Binti Mohd YusofClass : 5 BijakTeacher : Mrs. Haniza

Acknowledgement……………..………………………….……………3

Objective……………………………..…………………….……………..4

Introduction…………………………..…………………….……………..5

Part A……………………………………………………….……………..14

Part B…………………………………………………………..…………..19

Part C……………………………………………………….……………..21

Further Exploration……………………………………….……………..25

Reflection………………………………………………….……………..33

Conclusion……………………………………………………………….34

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First of all, I would like to say Alhamdulillah, thanks to God for giving me the strength and health to do this project work successfully and finish it on time.

Not forgotten, thanks to my parents for providing everything, such as laptop and internet connection, which really was a big help to me in finishing this project up as I could surf the net to find information and guidance for me to make it work. Their advices, which I really needed to motivate myself in completing this project. They also had been supporting and encouraging me to complete this task soon as I could so that I would not procrastinate in doing so.

Then, I would like to thank my Additional Mathematics teacher, Mrs Hanizah, who had been the one to guide me and the whole class throughout this project. Even though I had some difficulties in finishing this task, I managed to finish it well as she taught me patiently till I got her point.

I want to thank the media, for providing me countless informations on this topic. I would be lost if there’s no internet at my home.

Thanks to my friends who had been always supporting me. Even though this project had to be done individually, we discussed with each other on anything that was related to this project via Twitter and text messages. We shared ideas and methods to answer those asked questions correctly.

Last but not least, thanks to anyone who had been contributed either directly or indirectly in completing this project work. Without them, I believed, this project work could not be done in such a good way.

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The aims of carrying out this project work are:

To apply and adapt a variety of problem-solving strategies to solve

problems.

To improve thinking skills.

To promote effective mathematical communication.

To develop mathematical knowledge through problem solving in a

way that increases students’ interest and confidence.

To use the language of mathematics to express mathematical

ideas precisely.

To provide learning environment that stimulates and enhances

effective learning.

To develop positive attitude towards mathematics.

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History of Equations.

It is often claimed that the Babylonians (about 400 BC) were the first to solve quadratic equations. This is an over simplification, for the Babylonians had no notion of 'equation'. What they did develop was an algorithmic approach to solving problems which, in our terminology, would give rise to a quadratic equation. The method is essentially one of completing the square. However all Babylonian problems had answers which were positive (more accurately unsigned) quantities since the usual answer was a length.

In about 300 BC Euclid developed a geometrical approach which, although later mathematicians used it to solve quadratic equations, amounted to finding a length which in our notation was the root of a quadratic equation. Euclid had no notion of equation, coefficients etc. but worked with purely geometrical quantities.Hindu mathematicians took the Babylonian methods further so that Brahmagupta (598-665 AD) gives an, almost modern, method which admits negative quantities. He also used abbreviations for the unknown, usually the initial letter of a colour was used, and sometimes several different unknowns occur in a single problem.

The Arabs did not know about the advances of the Hindus so they had neither negative quantities nor abbreviations for their unknowns. However al-Khwarizmi (c 800) gave a classification of different types of quadratics (although only numerical examples of each). The different types arise since al-Khwarizmi had no zero or negatives. He has six chapters each devoted to a different type of equation, the equations being made up of three types of quantities namely: roots, squares of roots and numbers i.e. x, x2 and numbers.

1. Squares equal to roots.2. Squares equal to numbers.3. Roots equal to numbers.4. Squares and roots equal to numbers, e.g. x2 + 10x = 39.5. Squares and numbers equal to roots, e.g. x2 + 21 = 10x.6. Roots and numbers equal to squares, e.g. 3x + 4 = x2.

Al-Khwarizmi gives the rule for solving each type of equation, essentially the familiar quadratic formula given for a numerical example in each case, and then a proof for each example which is a geometricalcompleting the square.

Abraham bar Hiyya Ha-Nasi, often known by the Latin name Savasorda, is famed for his book Liber embadorum published in 1145 which is the first book published in Europe to give the complete solution of the quadratic equation.

A new phase of mathematics began in Italy around 1500. In 1494 the first edition of Summa de arithmetica, geometrica, proportioni et proportionalita, now known as the Suma, appeared. It was written by Luca Pacioli although it is quite hard to find the author's name on the book, Fra Luca appearing in small print but not on the title page. In many ways the book is more a summary of knowledge at the time and makes no major advances. The notation and setting out of calculations is almost modern in style:

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6.p.R.10

18.m.R.90

____________________________

108.m.R.3240.p.R.3240.m.R.90

hoc est 78.

In our notation

(6 + √10) (18 - √90) = (108-√3240 + √3240 - √900)which is 78.

The last term in the answer 90 is an early misprint and should be 900 but the margin was too narrow so the printer missed out the final 0!

Pacioli does not discuss cubic equations but does discuss quartics. He says that, in our notation, x4 = a + bx2 can be solved by quadratic methods but x4 + ax2 = b and x4 + a = bx2

are impossible at the present state of science.

Scipione dal Ferro (1465-1526) held the Chair of Arithmetic and Geometry at the University of Bologna and certainly must have met Pacioli who lectured at Bologna in 1501-2. Dal Ferro is credited with solving cubic equations algebraically but the picture is somewhat more complicated. The problem was to find the roots by adding, subtracting, multiplying, dividing and taking roots of expressions in the coefficients. We believe that dal Ferro could only solve cubic equation of the form x3 + mx = n. In fact this is all that is required.

For, given the general cubic y3 - by2 + cy - d = 0, put y = x + b/3 to get x3 + mx = n where m = c - b2/3, n = d - bc/3 + 2b3/27.

However, without the Hindu's knowledge of negative numbers, dal Ferro would not have been able to use his solution of the one case to solve all cubic equations. Remarkably, dal Ferro solved this cubic equation around 1515 but kept his work a complete secret until just before his death, in 1526, when he revealed his method to his student Antonio Fior.

Fior was a mediocre mathematician and far less good at keeping secrets than dal Ferro. Soon rumours started to circulate in Bologna that the cubic equation had been solved.

Nicolo of Brescia, known as Tartaglia meaning 'the stammerer', prompted by the rumours managed to solve equations of the form x3 + mx2 = n and made no secret of his discovery.Fior challenged Tartaglia to a public contest: the rules being that each gave the other 30 problems with 40 or 50 days in which to solve them, the winner being the one to solve most but a small prize was also offered for each problem.

 Tartaglia solved all Fior's problems in the space of 2 hours, for all the problems Fior had set were of the form x3 + mx = n as he believed Tartaglia would be unable to solve this type. However only 8 days before the problems were to be collected, Tartaglia had found the general method for all types of cubics.

News of Tartaglia's victory reached Girolamo Cardan in Milan where he was preparing to publish Practica Arithmeticae (1539). Cardan invited Tartaglia to visit him and, after much

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persuasion, made him divulge the secret of his solution of the cubic equation. This Tartaglia did, having made Cardan promise to keep it secret until Tartaglia had published it himself. Cardan did not keep his promise. In 1545 he published Ars Magna the first Latin treatise on algebra.

Here, in modern notation, is Cardan's solution of x3 + mx = n.

Notice that (a - b)3 + 3ab(a - b) = a3 - b3

so if a and b satisfy 3ab = m and a3 - b3 = n then a - b is a solution of x3 + mx = n. 

But now b = m/3a so a3 - m3/27a3 = n, i.e. a6 - na3 - m3/27 = 0.

This is a quadratic equation in a3, so solve for a3 using the usual formula for a quadratic. Now a is found by taking cube roots and b can be found in a similar way (or using b=m/3a). Then x = a - b is the solution to the cubic.

Cardan noticed something strange when he applied his formula to certain cubics. When solving x3 = 15x + 4 he obtained an expression involving √-121. Cardan knew that you could not take the square root of a negative number yet he also knew that x = 4 was a solution to the equation. He wrote to Tartaglia on 4 August 1539 in an attempt to clear up the difficulty. Tartaglia certainly did not understand. In Ars Magna Cardan gives a calculation with 'complex numbers' to solve a similar problem but he really did not understand his own calculation which he says is as subtle as it is useless.

After Tartaglia had shown Cardan how to solve cubics, Cardan encouraged his own student, Lodovico Ferrari, to examine quartic equations. Ferrari managed to solve the quartic with perhaps the most elegant of all the methods that were found to solve this type of problem. Cardan published all 20 cases of quartic equations in Ars Magna. Here, again in modern notation, is Ferrari's solution of the case: x4 +px2 + qx + r = 0. First complete the square to obtain

x4 + 2px2 + p2 = px2 - qx - r + p2

i.e.(x2 + p)2 = px2 - qx - r + p2

Now the clever bit. For any y we have

(x2 + p + y)2 = px2 - qx - r + p2 + 2y(x2 + p) + y2

= (p + 2y)x2 - qx + (p2 - r + 2py + y2) (*)

Now the right hand side is a quadratic in x and we can choose y so that it is a perfect square. This is done by making the discriminant zero, in this case

(-q)2 -4(p + 2y)(p2 - r + 2py + y2) = 0.Rewrite this last equation as

(q2 - 4p3 + 4 pr) + (-16p2 + 8r)y - 20 py2 - 8y3 = 0

to see that it is a cubic in y.

Now we know how to solve cubics, so solve for y. With this value of y the right hand side of (*) is a perfect square so, taking the square root of both sides, we obtain a quadratic in x. Solve this quadratic and we have the required solution to the quartic equation.

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The irreducible case of the cubic, namely the case where Cardan's formula leads to the square root of negative numbers, was studied in detail by Rafael Bombelli in 1572 in his work Algebra.In the years after Cardan's Ars Magna many mathematicians contributed to the solution of cubic and quartic equations. Viète, Harriot, Tschirnhaus, Euler, Bezout and Descartes all devised methods.Tschirnhaus's methods were extended by the Swedish mathematician E S Bring near the end of the 18th Century.

Thomas Harriot made several contributions. One of the most elementary to us, yet showing a marked improvement in understanding, was the observation that if x = b, x = c, x = d are solutions of a cubic then the cubic is

(x - b)(x - c)(x - d) = 0.

Harriot also had a nice method for solving cubics. Consider the cubic

x3 + 3b2x = 2c3

Put x = (e2 - b2)/e. Thene6 - 2c3e3 = b6

which is a quadratic in e3, and so can be solved for e3 to get

e3 = c3 +√(b6 + c6).

However

e3(e3 - 2c3) = b6 so that b6/e3 = -c3 +√(b6 + c6).

Now x = e - b2/e and both e and b2/e are cube roots of expressions given above.

Leibniz wrote a letter to Huygens in March 1673. In it he made many contributions to the understanding of cubic equations. Perhaps the most striking is a direct verification of the Cardan-Tartaglia formula. This Leibniz did by reconstructing the cubic from its three roots (as given by the formula) as Harriot claimed in general. Nobody before Leibniz seems to have thought of this direct method of verification. It was the first true algebraic proof of the formula, all previous proofs being geometrical in nature.

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Application of Equations.

We now fastforward 1000 years to the Ancient Greeks and see what they made of quadratic equations. The Greeks were superb mathematicians and discovered much of the mathematics we still use today. One of the equations they were interested in solving was the (simple) quadratic equation:

     

They knew that this equation had a solution. In fact it is the length of the hypotenuse of a right angled triangle which had sides of length one.

It follows from Pythagoras’ theorem that if a right-angled triangle has shorter sides   and   and hypotenuse  then

     

Putting   and   then  . Thus 

So, what is   in this case? Or, to ask the question that the Greeks asked, what sort of number is it? The reason that this mattered lay in the Greek’s sense of proportion. They believed that all numbers were in proportion with each other. To be precise, this meant that

all numbers were fractions of the form   where   and   are whole numbers. Numbers like

1/2, 3/4 and 355/113 are all examples of fractions. It was natural to expect that  was also a fraction. The huge surprise was that it isn’t. In fact

     

where the dots   mean that the decimal expansion of   continues to infinity without any discernible pattern. (We will meet this situation again later when we learn about chaos.)

 was the first irrational number (that is, a number which is not a fraction, or rational), to

be recognised as such. Other examples include  , ,   and in fact "most" numbers. It took

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until the 19th century before we had a good way of thinking about these numbers. The

discovery that   was not a rational number caused both great excitement (100 oxen were sacrificed as a result) and great shock, with the discoverer having to commit suicide. (Let this be an awful warning to the mathematically keen!) At this point the Greeks gave up algebra and turned to geometry.

Far from being an obscure number, we meet   regularly: whenever we use a piece of A4 paper. In Europe, paper sizes are measured in A sizes, with A0 being the largest with an area of  . The A sizes have a special relationship between them. If we now do a bit of origami, taking a sheet of A1 paper and then folding it in half (along its longest side), we get A2 paper. Folding it in half again gives A3, and again gives A4 etc. However, the paper is designed so that the proportions of each of the A sizes is the same - that is, each piece of paper has the same shape.

We can pose the question of what proportion this is. Start with a piece of paper with sides xand y with x the longest side. Now divide this in two to give another piece of paper with sidesy and x/2 with now y being the longest side. This is illustrated to the right.

The proportions of the first piece of paper are   and those of the second are   

or  . We want these two proportions to be equal. This means that

     

or

     

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Another quadratic equation! Fortunately it's one we have already met. Solving it we find that

     

This result is easy for you to check. Just take a sheet of A4 (or A3 or A5) paper and measure the sides. We can also work out the size of each sheet. The area   of a piece of A0 paper is given by

     

But we know that   so we have another quadratic equation for the longest side   of A0, given by

     

This means that the longest side of A  is given by   (why?) and that of A  

by  . Check these on your own sheets of paper.Paper used in the United States, called foolscap, has a different proportion. To see why, we return to the Greeks and another quadratic equation. Having caused such grief, the quadratic equation redeems itself in the search for the perfect proportions: a search that continues today in the design of film sets, and can be seen in many aspects of nature.

Let’s start with a rectangle, and then remove a square from it with the same side length as the shortest side of the rectangle. If the longest side of the rectangle has length 1 and the shortest side has length  , then the square has sides of length  . Removing it from the rectangle gives a smaller rectangle with longest side   and smallest side  . So far, so abstract. However, the Greeks believed that the rectangle which had the most aesthetic proportions (the so called Golden Rectangle) was that for which the large and the small rectangles constructed above have the same proportions. For this to be possible we must have

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This is yet another quadratic equation: a very important one that comes up in all sort of applications. It has the (positive) solution

     

The number   is called the golden ratio and is often denoted by the Greek letter  .

The Golden Rectangle can be seen in the shape of windows, especially on Georgian houses. More recently, the Golden Ratio can also be found as the "perfect shape" for photographs and film images. The quadratic equation   also arises in studies of the populations of rabbits and in the pattern in which the seeds of sunflowers and the leaves on the stems of plants are arranged. These are all linked with the Golden ratio through the Fibonacci sequence which is given by

     

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Sunflower seeds, arranged using Fibonacci numbers

The Parthenon, embodying the Golden Ratio

In this sequence each term is the sum of the previous two terms. Fibonacci discovered it in the 15th century in an attempt to predict the future population of rabbits. If you take the ratio of each term to the one after it, you get the sequence of numbers

     

and these numbers get closer and closer to (you guessed it) the Golden Ratio  .

By finding both of the roots of the above quadratic equation we can actually find a formula for the nth term in the Fibonacci sequence. If   is the  th such number with   and   then  is given by the formula

     

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PART A.

(a) Equation 1 : Axis of symmetry, x = 0.

Method 1: General Form ,with c = 175

General Form

Passing through (50,100),

Passing through (-50,100),

Quadratic Equation

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(0, 175)

(50, 100) (-50, 100)

y

x0

Equation 2 : Axis of symmetry, x = 50.

Method 1: General Form, with c =100

General Form

Passing through (50,175),

Passing through (100,100),

Quadratic Equation

Equation 3 : Axis of symmetry, x = 0.

Method 1: General Form, with c =75

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0 x

y

(100, 100)(0, 100)

(50, 175)

y

x0 (0, 75)

(50, 0)(-50, 0)

General Form

Passing through (-50,0) ,

Passing through (50,0) ,

Quadratic Equation

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(b) Method 1

Area of region A =

Area of region B = 100 x 100= 10000 cm2

Total surface area = 10 000 + 5 000 = 15 000 cm2

Method 2

(Refer to Equation 2)

Area =

Region A

Region B

(Refer to Equation 3)

PART B.

(a) Structure 1

Area ,

Total Area = 1x1 + 0.5= 1.5 m3

Total Volume = 1.5 x 0.13 = 0.195 m3

Cost = 0.195 x RM 960 = RM 187.20

Structure 2

Area

= 0.1875 m2

Total Area = 0.1875 + 1 m2

= 1.1875 m2

Total Volume = 1.1875 x 0.13 = 0.154375 m3

Cost = 0.154375 x RM 960 = RM 148.20

Structure 3

Area

= 0.5625 m2

Total Area = 0.1875 + 1 m2

= 1.5625 m2

Total Volume = 1.5625 x 0.13 = 0.203125 m3

Cost = 0.203125 x RM 960 = RM 195.00

Structure 4

eArea

= 0.46875 m2

Total Area = 0.46875+ 1 m2

= 1.46875 m2

Total Volume = 1.46875 x 0.13 = 0.190375 m3

Cost = 0.190375 x RM 960 = RM 183.30

Answer: Structure 2

(b) Structure 2. It consumes the lowest cost.

PART C.

(a)

Area of triangle ACE

Area of triangle ABD

Area of triangle BCF

cm2

(b)

Y = , X = , m = , c =

1 2 3 4 568.42 135.10 200.05 263.27 324.76

68.42 67.55 66.68 65.82 64.95

*Refer the graph paper next page.

From the graph,

(c)

Method 1: Differentiation

Method 2: Completing the square

Conics link quadratic equations to the stars

The Greeks were also very interested in the shape of cones. The picture above shows a typical cone.

Half of the cone can be visualised as the spread of light coming from a torch. Now, if you shine a torch onto a flat surface such as a wall then you will see various shapes as you move the torch around. These shapes are called conic sections and are the curves that you obtain if you take a slice through a cone at various different angles. Precisely these curves were studied by the Greeks, and they recognised that there were basically four types of conic section. If you take a horizontal section through the cone then you get a circle. A section at a small angle to the horizontal gives you an ellipse. If you take a vertical section then you get a hyperbola and if you take a section parallel to one side of the cone then you get a parabola. These curves are illustrated below.

A cross-section of a cone can be a circle ...

... an ellipse ...... a parabola ...

... or a hyperbola.

Conic sections come into our story because each of them is described by a quadratic

equation. In particular, if   represents a point on each curve, then a quadratic equation links   and  . We have:

The circle:  ;

The ellipse:  ;

The hyperbola:  ;

The parabola: 

These curves were known and studied since the Greeks, but apart from the circle they did not seem to have any practical application. However, as we shall see in the next issue of Plus, a link between quadratic equations and conics, coupled with a mighty lucky fluke, led to an understanding of the way that the universe worked, and in the 16th century the time came for conics to change the world.

Problem solving involving quadratic equations.

In the following examples, I will show how a word problem can be solved using a quadratic equation.

Solving Right Triangles Using the Pythagorean Theorem.

Recall that for a right triangle the sum of the squares of the legs is equal to the square of the Hypotenuse:

Example 1:

Suppose that one leg of a right triangle is 12 inches while the hypotenuse is   inches.. Find the length of the other leg.

Solution

Let x be the length of the other leg. Substituting into the Pythagorean theorem we have

.

Since the right side is equal to  This equation simplifies to

or  .

This means that x is 4 or -4.

Only 4 can be a length of the side of a triangle; so the other leg is 4 inches long.

Example 2

Suppose that one leg of a right triangle is 1 more than the other leg; and the hypotenuse is 1 less than 2 times the shorter leg. Find the lengths of all the sides.

Solution

Let one side have length x. then the other side can be expressed as x + 1 (the longer leg). The hypotenuse would then be expressed as 2x - 1. So by the Pythagorean theorem we have the equation

 .

We need to solve this equation for x. We must first expand (multiply) all of the terms:

so we have the equation

.

Combining like terms and setting one side to zero we have

.

Factoring the right hand side gives

If we set each of these factors equal to zero we get two solutions: x = 0 or x = 3. Only 3 can be the length of a side of a triangle. So the lengths are : 3, 4, and 5.

Number problems.

Recall that if x is an integer, then the next consecutive integer is x + 1. For example, the next consecutive integer after 7 is 7 +1 = 8.

If x is an even integer the next consecutive even integer would be x + 2. For example, the next consecutive even integer after -6 is -6 + 2 = -4.

If x is and odd integer the next consecutive odd integer would be x + 2 as well.   For example, the next consecutive odd integer after -9 is -9 + 2 = -7.

Example 3:Find two positive consecutive odd integers whose product is 99.

Solution

Let x be the first integer. Then the next odd integer is x + 2. So we have

x(x+2) = 99.

To solve this equation first we distribute and then set one side to zero. We have

Factoring the left side gives

(x + 11)(x - 9) = 0.

So the solutions are x = -11 or x = 9. Since we are looking for positive integers, the answers are 9 and 11.

Area Problems.

Recall the following formulas for a rectangle:

Perimeter:  P = 2L + 2 WArea:   A = LW

Example 4

The width of a rectangle is 16 feet less than 3 times the length. If the area is 35 square feet, find the dimensions of the rectangle.

Solution

Let x be the length because the width is expressed in terms of the length. So the length is 3x - 16. The total area is 35 square feet so we have the equation

35 = x(3x -16).

To solve we first distribute:

                  35 = 3x2  - 16x

then set the left side to zero:

.

Factoring gives

0 = (3x +5)(x -7).

Only the second factor will give a positive solution, so the answer is 7. The dimensions of the rectangle are: Length: 7 feet width: 5 feet.

Finding areas by integration

Integration can be used to calculate areas. In simple cases, the area is given by a single definiteintegral. But sometimes the integral gives a negative answer which is minus the area, and inmore complicated cases the correct answer can be obtained only by splitting the area into several parts and adding or subtracting the appropriate integrals.

Whenever we are calculating area in a given interval, we are using definite integration. Let’s try to find the area under a function for a given interval.

(1) Integrate   from [-2, 2].

Step 1: Set up the integral.

Step 2: Find the Integral.

*Note: We don't have to add a "+C" at the end because it will cancel out finding the area anyway.

Step 3: Integrate from the given interval, [-2,2].

The area of the curve to the x axis from -2 to 2 is 32⁄3 units squared.

On the graph, the red below the parabola is the area and the dotted line is the integral function. Notice that the integral function is cubic and the original function is quadratic. The integral will always be a degree higher than the original function. Looking at the graph, there is a geometric relationship between the original function and the integral function. We can see at x = -2 the integral function has a y value of a little under -5, and at x = 2 the integral has a y value of a little over 5. The difference of 5.3 and -5.3 gives us an area of 32⁄3, which is a little over 10.

When taking the definite integral over an interval, sometimes we will get negative area because the graph interprets area above the x axis as positive area and below the x axis as negative area.

After spending countless hours, days and night to finish this project and also sacrificing my time for video games and stuffs during this mid-year school break, there are several things that I can say. I’m going to express it through words anyways.

After doing some researches, answering the given questions, drawing the graphs and some problem solving, I saw that the usage of quadratic

equation is important in daily life. It is not just widely used in architecture such as determining the area of a sculpture with curve(s) but we use

quadratic equation in our daily life as well. To be related, determining the area is important as it can give the exact amount of the needed cost.

But, what is the use of quadratic equation in daily life of normal people like us? In reality most people are not going to use the quadratic equation in

daily life. Having a firm understanding of the quadratic equation as with most maths helps increasing logical thinking, critical thinking, and number sense.

We use quadratic equations to determine how to shape the mirror of, say a car headlight, that is familiar, and where to put the light. If the light is at the

focus, as it should be, all light from the bulb will be reflected straight out.

As a conclusion, quadratic equation is a daily life essentiality. If there is no quadratic equation, architect won’t be able to create such perfect buildings,

and light from bulbs in front of a car cannot shine brilliantly.