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In the present day, we are entitled to discuss something out of Mathematics, the Quadratic Equation. Although, this is a famous subject matter amidst different exams and studies, it requires sufficient amount of practice, subsequently. Students often get confused and relate quadratic equations with linear equation. Let me make it clear, my patrons, it is entirely distinct from the Linear Equations.

## Name and Origin of the Quadratic Equation:

To the best of our knowledge the origin of the term ’Quadratic’ is Latin.

It is derived from quadraters which the past participle of ‘Quadrare’.The name ‘Quadratic’ comes from ‘Quad’ meaning square, because the variable gets squared (like x2). It is also called an ‘Equation of Degree’ (because of the 2 on the x).

The main difference among these equations is the latter will definitely comprise of a term ‘x2’. The important fact to keep in mind is, a term with power 2 or a term with a degree 2 in an equation makes it a quadratic equation. This article further discusses the significance of quadratic equations and its value in the colossal of mathematics.

## Analysis

A quadratic formula is significant to resolve a quadratic equation, in elementary algebra. Even though, there are various other methods to solve the quadratic equation, for instance graphing, completing the square, or factoring; yet again, the most convenient and easy approach to work out these quadratic equations is the quadratic formula.

The most common form of a quadratic equation is:

$$a x^2 + bx + c = 0$$

Here a, b and c are constants and x denote to be undetermined with a not equal to 0. A quadratic formula thereby, satisfies a quadratic equation by placing the former into the latter. Considering the above parameters, the quadratic formula may signify as,

$$x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a }$$

It is important to learn by heart, that every single solution conferred through the quadratic formula is considered to be a derivation of the quadratic equation. According to the rules of Geometry, this derivation certainly denotes the values for ‘x’ where any parabola clearly denoted as y = ax2 + bx + c intersects the x-axis. This formula engenders the zeros of any parabola and further yields the axis of symmetry of that parabola. Moreover, this formula is essentially used to analyze the number of real zeros present in the quadratic equation.

## Derivation of the Quadratic Formula:

We know that the standard form of Quadratic Equation with rational coefficient is:

$$a x^2 + bx + c \ne 0$$

Divide the whole equation by ‘a’:

$$x^2 + \frac { bx } { a } + \frac { c } { a } = 0$$ ……….. this is equation #1

Shift constant term to other side:

$$x^2 + \frac { bx } { a } = – \frac { c } { a }$$ ……….. this is equation #2

To make left hand side a perfect square, add $$[ \frac {1}{2} * \frac {b}{2} ]^2 = \frac{b^2}{4a^2}$$ to both the side of equation number 2.

Taking square root of both sides:

$$\sqrt { (x + \frac {b} {2a} )^2 } = \sqrt { \frac {b^2 – 4ac} {4a^2} }$$

$$=> x + \frac{b}{2a} = \pm \sqrt { \frac {b^2 – 4ac} {2a} }$$

$$=> x = – \frac {b}{2a} \pm \sqrt { \frac {b^2 – 4ac} {2a} }$$

$$=> x = \frac {-b \pm \sqrt {b^2 – 4ac} } {2a}$$

## Discriminant:

We call the term $$b^2 – 4ac$$ the Discriminant form.The Discriminant is important.It tells how many quadratic root a quadratic function has:

1. $$b^2 – 4ac < 0$$ : There is no equal root.
2. $$b^2 – 4ac = 0$$ : There is one real root.
3. $$b^2 – 4ac > 0$$ : There is two real root.

Let us discuss here the Quadratic Equation Solver:

It is often admitted by the mathematicians though, that the quadratic formula is chaotic and prodigious, at times. Make sure, you are not confused or distracted from the persona of this very important equation going through a horde of numbers, letters and signs.

## Working Out the Quadratic Formula:

Quadratic equations are significantly solved via quadratic formula, and is considered among the top five formulae in the subject of mathematics. It is not important for the students to learn the formulas thoroughly and commit to memory. However, this one is quite resourceful and it is recommended to learn it by heart, not only how to derive it but also how to make use of it.

## Solving a Problem Using the Quadratic Formula

The first step to be followed, should be to recognize the values for a, b and c, that are said to be the coefficients. Be confident, and verify if the given equation is in alliance with the basic quadratic formula, i.e., $$a x^2 + bx + c = 0$$:

Remember that ‘a’ is never equal to zero, owing to the significance of x2 it becomes a quadratic.

Let’s find the solution for this equation. $$x^2 + 5x + 6 = 0$$ .

• Since, ‘a’ is said to be the coefficient against x2, for this reason, a=1. ‘b’, on the contrary, stands in front of ‘x’ as a coefficient, which makes $$b = -5$$.
• A constant is something that stands individual in the equation and do not hold any ‘x’ alongside. For this reason, here $$c = 6$$.

$$x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a }$$

$$x = \dfrac{ -(-5) \pm \sqrt{(-5)^2 – 4(1)(6)}}{ 2(1) }$$

solving this looks like:

$$x = \dfrac{ 5 \pm \sqrt{25 – 24}}{2}$$

$$x = \dfrac{ 5 \pm \sqrt{1}}{2}$$

$$x1 = \dfrac{ 5 + 1 }{2}$$

$$x2 = \dfrac{ 5 – 1 }{2}$$

The discriminant $$b^2 – 4ac > 0$$

There are two equal roots,so the answer is:

$$x1 =3.000$$

$$x2 = 2.000$$

## Solving Quadratic Equation by Graphing method:

From the graph of the parabola determine the vertex,axis of symmetry,y-intercept,x-intercept.

The problem has two solutions and they both demonstrate the intersecting points of the equation, that is, the x – intercept, the point where the x-axis is crisscrossed by a curve. Whilst, preparing a graph the equation $$x^2 + 3 x – 4 = 0$$, can be viewed as:

1) Vertex: It means ‘peak’ .So the vertex of quadratic equation means peak point of parabola.If the parabola opens upward, the vertex is the highest point and if the parabola opens downward ,so the vertex is the lowest point.

2) Axis of symmetry: The axis of symmetry always passes through the vertex of parabola.It divides the parabola into two equal halves.

3) X-intercept: Roots are also called x-intercept.In the graph it is located below the x-axis or above the x-axis.Therefore to find the root of a quadratic function,we set y=0.

4) Y-intercept: Every parabola contains y-intercept,the point at which the function crosses the y-axis.It is found by setting the x-variable in the equation to 0.

Let’s solve graphically.

Take the equation $$f(x)= 2^2 – 4x – 1$$ or $$Y=2^2 – 4x – 1$$

Since, $$a=1$$, $$b=-4$$ and $$c=-1$$.

If “a” is having the positive value ,the parabola opens upward in graph.First we find the vertex of x:

$$x = \frac{-b}{2a}$$

$$x = \frac{-(-4)}{2(2)}$$

$$x = 1$$

Then we find the vertex of Y:

We put the value of x in equation $$2^2 + 4 x – 1 = 0$$

$$y = 2(1)^2 – 4(1) – (1)$$

$$y = 2 – 4 – 1$$

$$y = -3$$

We have axis of symmetry: $$x = 1$$

Now,we find x-intercept.To find x-intercept we use quadratic formula:

$$x = \dfrac{ -(-4) \pm \sqrt{(-4)^2 – 4(2)(-1)}}{ 2(2) }$$

$$x = \dfrac{ 4 \pm \sqrt{16 + 8}}{ 4) }$$

$$x = \dfrac{ 4 \pm \sqrt{24}}{4) }$$

$$x = \dfrac{ 4 \pm \sqrt{4.9}}{ 4) }$$

$$x = \dfrac{ 4 + \sqrt{4.9}}{ 4) }$$ , $$x = \dfrac{ 4 – \sqrt{4.9}}{ 4) }$$

$$x-intercept = 2.23, -0.23$$

Now, we find y-intercept, we put the value of x = 0 in equation as:

$$y = 2^2 – 4x – 1$$

$$y = 2(0)^2 – 4(0) – 1$$

$$y-intercept = -1$$

now ,we plot the values in graph. Let’s take another equation in which parabola opens downward.

$$-x^2 + 2x + 1 = 0$$

* if ‘a’ is having negative value so the parabola opens downward.

Find vertex of x:

$$x = \frac{-b}{2a}$$

$$x = \frac{-(2)}{2(-1)} = 1$$

Find vertex of y:

We put the value of x in equation,

$$y = -(1)^2 + 2(1) + 1$$

$$y = 2$$

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

$$a = -1$$, $$b = 2$$, $$c = 1$$

$$x = \frac{-2 \pm \sqrt{2^2 – 4(-1)(1)}}{2(-1)}$$

$$x = \frac{-2 \pm \sqrt{8}}{-2}$$

$$x1 = -0.414214$$

$$x2 = 2.414214$$

Now,find y-intercept

$$x^2 +2x + 1 = 0$$

$$(0)^2 +2(0) + 1 = 0$$

y-intercept = 1. Now,we plot the values in graph: ## Quadratic equation by Factoring Method:

Factorize the left side of Quadratic equation and then write the product of these two factors equal to zero.Equate each factor to zero and solve for ‘x’ .

Here we have $$x^2 +5x + 6 = 0$$

It gives $$a = 1$$, $$b = -5$$, $$c = 6$$

Now,we multiply ‘a’ by ‘c’ i.e. 1 * 6

Then make two such factors of ‘ac=6,whose product equals ‘ac=6’ but its sum equals ‘b=-5’.

Therefore ‘-3’ and ‘-2’ are such factors where product is 6 and whose sum is -5.Then we can write,

$$x^2 – 3x – 2x + 6 = 0$$

$$x(x-) – 2(x-3) = 0$$

Either $$(x-3) = 0$$ then $$x = 3$$,

Or $$(x-2) = 0$$ then $$x = 2$$

Hence the solution is:

$$x1 = 3.000$$

$$x2 = 2.000$$

As a student you might need it in various topics regarding mathematics. Also, you shall make use of this equation in subjects like engineering and physics. There are a few steps, if followed pertinently, the factor of being distracted for the students might lessen to some extent.

## Professions that use Quadratic Equation:

### 1) Military and Law Enforcement:

The Quadratic Equations is used by the military to find the trajectory of missiles fired by artillery. It is often used to describe the motion of objects that fly through the air.People who join the military and work with artillery or tanks use quadratic equation to predict where shells will land.Police also use it to determine the trajectories of bullets. When you traject a missile (or shoot an arrow,throw a ball or a stone) it goes up into the air,slowly as it travels,then comes down again faster and faster,and the quadratic equation tells you its position at all times!

Example: Traject a missile.

A missile is trajected, from 10m above the ground with a velocity of 14m/s.When does it fall on the ground?

Ignoring air resistance,we can work out its height by adding these three things.

The height starts at 10 meter=10.

It travels upwards at 14m per second = $$14m/s = 14t$$.

Gravity pull it down,changing its position by about $$-5m/s^2 = -5t^2$$ .

Add them up and the ’h’ denotes height and the ‘t’ denotes time in equation.

And the missile will fall on the ground when the height is zero.

$$10 + 14t – 5t^2 = 0$$

In standard form it looks like:

$$5t^2 + 14t + 10 = 0$$

Let’s solve the equation,

$$x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a }$$

$$x = \dfrac{ -14 \pm \sqrt{14^2 – 4(-5)(10)}}{ 2(-5) }$$

$$x = \dfrac{ -14 \pm \sqrt{196 + 200}}{ -10 }$$

$$x = \dfrac{ -14 \pm \sqrt{396}}{ -10 }$$

The discriminant  $$b^2 – 4ac > 0$$

There are two equal roots:

$$x1 = -0.58$$

$$x2 = 3.38$$

### 2) ENGINEER: It is necessary for the ENGINEERS to design of any piece of equipment that is curved.Like bridges, roller coaster rides, auto bodies. Automotive engineers also use them to design brake system. Audio engineers use these equations to design sound system that have the best sound quality possible.

The following example relates to the civil engineering.The cable joining the two towers of the Golden Gate bridge is modeled by $$y = 2x^2 – 8x + 12$$, where y is the height of the cables from the deck and X is the horizontal distance from the left pillar, where the height is equal to 16m.

To determine the result solve the equation: $$2x^2 – 8x + 12 = 16$$

$$2x^2 – 8x + 12 – 16 = 0$$

$$2x^2 – 8x – 4 = 0$$

We take ‘2’ common:

$$2 (x^2 – 4x – 2) = 0$$

Where  $$a = 1$$, $$b = -4$$, $$c = -2$$

now, we put the values in quadratic formula: $$x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a }$$

$$x = \dfrac{ -(-4) \pm \sqrt{(-4)^2 – 4(1)(-2)}}{ 2(1) }$$

$$x1 = \dfrac{ 4 + \sqrt{16 + 8}}{ 2 }$$ and $$x2 = \dfrac{ 4 – \sqrt{16 + 8}}{ 2 }$$

$$x1 = 2 + 6$$ and $$x2 = 2 – 6$$

At the point that are $$x1 = 2 + 6$$ and $$x2 = 2 – 6$$  m away from the left pillar,the height of the hanging cable is 16m.

### 3) Equation in motion:

Quadratic functions are used in playground as well and in gaming situations.It’s describing the trajectory of a ball and determining the height of a thrown ball. ### 4) Science:

Astronomers use Quadratic Equations to describe the orbit of planets,solar systems and galaxies.

### 5) Agriculture sectors:

Quadratic Equation is  also used in agriculture sectors.One of the use is to find out the optimal arrangement of boundaries to produce the biggest field.Area is the length of a surface multiplied by its width.This turns calculation involving area into the Quadratic Equations.