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 a sufficient amount of practice, subsequently. Students often get confused and relate quadratic equations with a linear equation. Let me make it clear, my patrons, it is entirely distinct from the Linear Equations. Are you Student? Learn more about Quadratic equations

# Quadratic Formula Calculator

Our quadratic equation solver will solve a second-order polynomial equation such as \( ax^2 + bx + c = 0 \) for \( x \), where \( a ≠ 0 \), using the quadratic formula.

The calculator solution will show steps using the quadratic formula to solve the entered equation for real and complex roots.

This calculator also determines whether the discriminant \( b^2 – 4ac \) is less than, greater than or equal to 0.

When \( b^2 – 4ac < 0 \) there is no equal root.

When \( b^2 – 4ac = 0 \) there is one real root.

When \( b^2 – 4ac > 0 \) there is two real root.

## Quadratic Formula:

\( 2 x^2 + -12x + 7 = 0 \)

where \( a ≠ 0 \) . (second order polynomial), Quadratic formula is used to solve quadratic equation:

## Quadratic Equation:

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

## Examples using quadratic formula:

**Example 1:** Let’s find the solution for \( a x^2 + bx + c = 0 \), where \( a = 2 \), \( b = -12 \), \( c = 7 \), using the Quadratic Formula:

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

\( x = \dfrac{ -(-12) \pm \sqrt{(-12)^2 – 4(2)(7)}}{ 2(2) } \)

\( x = \dfrac{ 12 \pm \sqrt{144 – 56}}{4} \)

\( x = \dfrac{ 12 \pm \sqrt{88}}{4} \)

The discriminant \( b^2 – 4ac > 0 \), so there are two roots:

**simplify the radical**

\( x = \dfrac{ 12 \pm 2\sqrt{22}\, }{ 4 } \)

\( x = \dfrac{ 12 }{ 4 } \pm \dfrac{2\sqrt{22}\, }{ 4 } \)

Simplify fractions and/or signs:

\( x = 3 \pm \sqrt{22}\, \)

which becomes

\( x = 5.34521 \)

\( x = 0.654792 \)

**Example 2:** Let’s find the solution for \( a x^2 + bx + c = 0 \), where \( a = 3 \), \( b = 14 \), \( c = 18 \), using the Quadratic Formula:

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

\( x = \dfrac{ -(14) \pm \sqrt{(14)^2 – 4(3)(18)}}{ 2(3) } \)

\( x = \dfrac{ -14 \pm \sqrt{196 – 216}}{6} \)

\( x = \dfrac{ -14 \pm \sqrt{-20}}{6} \)

The discriminant \( b^2 – 4ac > 0 \), so there are two roots:

**simplify the radical**

\( x = \dfrac{ -14 \pm 4\sqrt{15}\, i }{ 6 } \)

\( x = \dfrac{ -14 }{ 6 } \pm \dfrac{2\sqrt{5}\, i}{ 6 } \)

Simplify fractions and/or signs:

\( x = \dfrac{ -7 }{ 3 } \pm \dfrac{2\sqrt{5}\, i}{ 3 } \)

which becomes

\( x = -2.33333 + 0.745356 \, i \)

\( x = -2.33333 – 0.745356 \, i \)

## 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 quadratus which the past participle of ‘Quadrare’.The name ‘Quadratic’ comes from ‘Quad’ meaning square because the variable gets squared (like x^{2}). It is also called an ‘Equation of Degree’ (because of the 2 on the x).

## An important fact about the Quadratic Equation:

The main difference among these equations is the latter will definitely comprise of a term ‘x^{2}’. 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 their value in the colossal of mathematics.

**Where Can You Use The Quadratic Equation Formula?**

You can calculate the expected profit you are going to get if you are using the quadratic equation. It can prevent unwanted surprises and provide you with the accurate numbers and what to expect in the future. Even in the business where you are simply selling bottled water, it can help you estimate how many bottles you have to sell to generate the profit you want.

Whether you are buying new furniture or a new carpet for your room – you need to calculate the area of the rooms and that’s where quadratic formula equation comes to your rescue. With the incorporation of the variables, you can accurately calculate the area of rooms and revamp the looks of your room accordingly.

Quadratic formula equation is also useful when it comes to athletic activities especially throwing balls calculating the speed and distance. You can estimate the speed you will need to throw the ball at a specific distance because when it comes to sports, mathematical calculations can be your best friend.

**From Basic To Technicalities:**

Now that we have understood the practical uses of the quadratic formula equation, let’s explore the basics of it.

A quadratic equation is a polynomial equation that means it includes different variables, coefficients, and exponents. These variables can be repeated in the equation and a graph can also be drawn utilizing intercepts. Quadratic Function Equation is the second-degree polynomial equation which is written in the following form

*ax ^{2} + bx + c = 0*

Here is the complete description of the variables.

- ‘a’ is the quadratic coefficient
- ‘x’ is the unknown
- ‘b’ is the linear coefficient
- ‘c’ is the constant

Here it is also important to note that the numerals i.e. a, b, and c are the coefficient of the equation and they cannot be ‘0’

Our quadratic formula calculator can help you if you can present the equation in this form

*ax ^{2} + bx + c = 0*

The solution of this equation is also known as the root of the equation.

**Calculations:**

The quadratic formula is as follows:

\( x = \dfrac{ -B \pm \sqrt{Δ}}{ 2A } \)

where

\( Δ = B^2 – 4AC\)

Using this formula, you can find the solutions of any quadratic equation. Note that there are three possible options of obtaining a result:

- The quadratic equation has two unique roots when Δ > 0. Then, the first solution of the quadratic formula is \( x₁ = \dfrac{ -B \pm \sqrt{Δ}}{ 2A } \) and the second is \( x₂ = \dfrac{ -B \pm \sqrt{Δ}}{ 2A } \).
- The quadratic equation has only one root when Δ = 0. The solution is equal to \( x = \dfrac{ -B }{ 2A } \). It is sometimes called a repeated or double root.
- The quadratic equation has no real solutions for Δ < 0.

You can also graph the function \( y = Ax^2 + Bx + C \). It’s shape is a parabola, and the roots of the quadratic equation are the x-intercepts of this function.

A, B and C are the coefficients of the quadratic equation. They are all real numbers not dependent on x. If \( A = 0 \), then the equation is not quadratic, but linear.

If \( B² < 4AC \), then the determinant Δ will be negative. It means that such an equation has no real roots.

**Step-by-step Solution**

- Write down your equation. Let’s assume it is \( 4x^2 + 3x – 7 = -4 – x \).
- Bring the equation to the form \( Ax^2 + Bx + C = 0 \). In this example, we will do it in the following steps:
- Calculate the determinant.\( Δ = B² – 4AC = 4² – 4*4*(-3) = 16 + 48 = 64 \).
- Decide whether the determinant is greater, equal or lower than 0. In our case, the determinant is greater than 0, what means that there are two unique roots of this equation.
- Calculate the two roots using the quadratic formula. \( x₁ = \dfrac{ -B + \sqrt{Δ}}{ 2A } = \dfrac{ -4 + \sqrt{64}}{ 2*4 } = \dfrac{ -4 + 8}{ 8 } = \dfrac{ 4 }{ 8 } = 0.5 \) and \( x₂ = \dfrac{ -B – \sqrt{Δ}}{ 2A } = \dfrac{ -4 – \sqrt{64}}{ 2*4 } = \dfrac{ -4 – 8}{ 8 } = \dfrac{ -12 }{ 8 } = -1.5 \)
- The roots of your equation are \( x₁ = 0.5 \) and \( x₂ = -1.5 \).

As the coefficients are described individually, you can enter the values accordingly and get the answer to your problem instantly. It can also help you to place the value on the graph and reflect the data in the graphical format so it can be easy to evaluate.

With the help of quadratic formula calculator, you can finally find the value of ‘x’ and solve the equation in seconds. Best of luck finding the ‘x’ so you can move to the next!

## 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:

- \( b^2 – 4ac < 0 \) : There is no equal root.
- \( b^2 – 4ac = 0 \) : There is one real root.
- \( b^2 – 4ac > 0 \) : 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 x
^{2}, 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 the quadratic equation means the peak point of the 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 a parabola. It divides the parabola into two equal halves.

3) **X-intercept**: Roots are also called the 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,

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

\( y = 2 \)

Find x-intercept by quadratic equation:

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

\( a = -1 \), \( b = 2 \), \( c = 1 \)

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

\( x = \frac{-2 \pm \sqrt{8}}{-2} \)

\( x1 = -0.414214 \)

\( x2 = 2.414214 \)

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:

\( x1 = 3.000 \)

\( x2 = 2.000 \)

## Importance of Quadratic Equation in Your Real Life.

As a student, you might need it on 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 the 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 the height and the ‘t’ denotes time in the equation.

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

\( 10 + 14t – 5t^2 = 0 \)

Which is the quadratic equation?

In standard form it looks like:

\( 5t^2 + 14t + 10 = 0 \)

Let’s solve the equation,

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

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

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

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 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.

Lessons for Quadratic Equations

In this section, you will learn about quadratic equation, how many forms of quadratic equations and methods to solve quadratic equations with examples.

## Standard form of Quadratic Equation

\( ax^2 + bx + c = 0 \)

This is the standard form of quadratic equations. In this form of quadratic equations we have,

- a, b and c are constants
- a is not equals to zero (\( a ≠ 0 \))

### For Example:

Lets suppose three equations A, B and C.

Equation A: \( x^2 + 5x + 2 = 0 \), here \( a = 1 \), \( b = 5 \) and \( c = 2 \)

Equation B: \( 3x^2 + x + 9 = 0 \), here \( a = 3 \), \( b = 1 \) and \( c = 9 \)

Equation C: \( x + 9 = 0 \), here \( a = 0 \), \( b = 1 \) and \( c = 9 \)

Equation A and B are standard form of quadratic equation, but equation C not belongs to standard form of quadratic because in this equation, \( a = 0 \)

## Hidden Quadratic Equations

As we know the standard form of quadratic equation is \( ax^2 + bx + c = 0 \), but sometime it looks different.

### For Example:

Lets suppose three equations A, B and C.

Equation A: \( 2x^2 = x – 1 \)

Equation B: \( 3(x^2 + 2x) = 10 \)

Equation C: \( x(2x + 1) = 5 \)

These equations doesn’t look like standard form of quadratic equations. Lets drive these equations in standard form:

Equation A: \( 2x^2 = x – 1 => 2x^2 – x + 1 = 0 \), here \( a = 2 \), \( b = -1 \) and \( c = 1 \)

Equation B: \( 3(x^2 + 2x) = 10 => 3x^2 + 6x = 10 => 3x^2 + 6x – 10 = 0 \), here \( a = 3 \), \( b = 6 \) and \( c = -10 \)

Equation C: \( x(2x + 1) = 5 => 2x^2 + x = 5 => 2x^2 + x – 5 = 0 \), here \( a = 2 \), \( b = 1 \) and \( c = -5 \)

Equation A, B and C are standard form of quadratic equation.

## Complete Quadratic Equations

When b is not equals to zero (\( b ≠ 0 \)), the equation is known as complete quadratic equation.

### For Example:

Equation A: \( x^2 + 5x + 2 = 0 \), here \( a = 1 \), \( b = 5 \) and \( c = 2 \)

Equation B: \( 3x^2 + x + 9 = 0 \), here \( a = 3 \), \( b = 1 \) and \( c = 9 \)

Equation C: \( x^2 + 9 = 0 \), here \( a = 1 \), \( b = 0 \) and \( c = 9 \)

Equation A and B are complete quadratic equation, whereas equation C not belongs to complete quadratic because in this equation, \( b = 0 \)

## Pure or Incomplete Quadratic Equations

When b is equals to zero (\( b = 0 \)), the equation is known as pure quadratic equation or incomplete quadratic equation.

### For Example:

Equation A: \( x^2 + 2 = 0 \), here \( a = 1 \), \( b = 0 \) and \( c = 2 \)

Equation B: \( 3x^2 + 9 = 0 \), here \( a = 3 \), \( b = 0 \) and \( c = 9 \)

Equation C: \( x^2 + x + 9 = 0 \), here \( a = 1 \), \( b = 1 \) and \( c = 9 \)

Equation A and B are incomplete quadratic equation, whereas equation C not belongs to incomplete quadratic because in this equation, \( b ≠ 0 \)

## Roots of Quadratic Equations

The roots in quadratic equation are values of x, roots are also called x-intercepts or zeros.

### For Example:

Lets suppose equation A: \( x^2 + 5x + 6 = 0 \). Find the roots of equation A.

\( => x^2 + 5x + 6 = 0 \)

\( => x^2 + 2x + 3x + 6 = 0 \)

\( => x(x + 2) + 3(x + 2) = 0 \)

\( => (x + 2)(x + 3) = 0 \)

\( => (x + 2) = 0, (x + 3) = 0 \)

\( => x + 2 = 0, x + 3 = 0 \)

\( => x = -2, x = -3 \)

Now replace x with -2 and -3 in equation A.

For \( x = -2 \), \( (-2)^2 + 5(-2) + 6 = 4 – 10 + 6 = 0 \)

For \( x = -3 \), \( (-3)^2 + 5(-3) + 6 = 9 – 15 + 6 = 0 \)

As you can see that by replacing x with -2 or -3 makes quadratic equation equals to zero, its mean -2 and -3 are roots of \( x^2 + 5x + 6 = 0 \)

## Solving Quadratic Equation

There are three methods to solve quadratic equations:

- Factorization
- Completing the square
- Quadratic Formula

## Solving Quadratic Equation by doing Factorization

### Steps of factorization:

- Convert the equation in standard form of quadratic equation which is \( ax^2 + bx + c = 0 \), make right hand side zero, if right hand side is not zero then take it to left hand side.
- Break the middle term of left hand side and factorize it.
- Break the equation in two equations depend on it factors and equate each equation to zero.

### For Example:

Lets suppose equation A: \( x^2 + 5x + 3 = -3 \). Find the roots of equation A.

**Step 1:**

\( => x^2 + 5x + 3 + 3 = 0 \)

\( => x^2 + 5x + 6 = 0 \)

**Step 2:**

\( => x^2 + (2 + 3)x + 6 = 0 \)

\( => x^2 + 2x + 3x + 6 = 0 \)

\( => x(x + 2) + 3(x + 2) = 0 \)

\( => (x + 2)(x + 3) = 0 \)

**Step 3:**

\( => (x + 2) = 0, (x + 3) = 0 \)

\( => x + 2 = 0, x + 3 = 0 \)

\( => x = -2, x = -3 \) roots of equation.

## Solving Quadratic Equation by completing square

In this type, we convert the equation in \( (x + p)^2 = q \) form.

### Steps of completing the square:

- Convert the equation in standard form of quadratic equation which is \( ax^2 + bx + c = 0 \), make right hand side zero, if right hand side is not zero then take it to left hand side.
- Divide both side by value of a (i.e. co-efficient of \( x^2 \)), and make \( a = 1 \).
- Shift the constant term(i.e. c) to right hand side.
- Convert left hand side in \( x^2 + 2(x)(\dfrac{b}{2}) + (\dfrac{b}{2})^2 = c + (\dfrac{b}{2})^2 \) form by adding square of \( \dfrac{b}{2} \) in both sides.
- As we know \( (a + b)^2 = a^2 + 2(a)(b) + (b)^2 \), convert left hand side in \( (a + b)^2 \) form and simplify right hand side.
- Solve equation by adding square root on both side.

### For Example:

Lets suppose equation A: \( -3x^2 + 12x + 3 = -2 \). Find the roots of equation A.

**Step 1:**

\( => -3x^2 + 12x + 3 + 2 = 0 \)

\( => -3x^2 + 12x + 5 = 0 \)

**Step 2:** Divide by -3 as a = -3

\( => \dfrac{(3x^2)}{-3} – \dfrac{12x}{-3} + \dfrac{5}{-3} = 0 \)

\( => x^2 – 4x – \dfrac{5}{3} = 0 \)

**Step 3:**

\( => x^2 – 4x = \dfrac{5}{3} \)

**Step 4:** Add \( 2^2 \) on both side as \( \dfrac{b}{2} = \dfrac{4}{2} = 2 \)

\( => x^2 – 4x + 2^2 = \dfrac{5}{3} + 2^2 \)

**Step 5:** \( (a – b)^2 = a^2 – 2(a)(b) + (b)^2 \), here we have a = x and b = 2

\( => (x – 2)^2 = \dfrac{5}{3} + 2^2 \)

\( => (x – 2)^2 = \dfrac{5}{3} + 4 \)

\( => (x – 2)^2 = \dfrac{5 + 12}{3} \)

\( => (x – 2)^2 = \dfrac{17}{3} \)

**Step 6:**

\( => \sqrt{(x – 2)^2} = \pm \sqrt{\dfrac{17}{3}} \)

\( => x – 2 = \pm \sqrt{\dfrac{17}{3}} \)

\( => x = 2 \pm \sqrt{\dfrac{17}{3}} \)

## Solve by using 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 x
^{2}, 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:

\( x1 =3.000 \)

\( x2 = 2.000 \)

You can also check our article: There Is More Than One Ways To Solve The Quadratic Equation

To calculate Discriminant Check discriminant calculator