## Basics

“Quadratic” word in quadratic equation comes from quadratic polynomial,
which is a polynomial with the highest degree of 2.
The term “Equation” is an expression
which
has sign of equality (“=”) between its left and right hand sides. So, it is written with
a quadratic polynomial on LHS and 0 on the RHS of the equation.

An example of a quadratic equation is 3x^{2} + 6x + 9 = 0. This example has 3x^{2} + 6x +
9 on LHS, which is a quadratic polynomial and 0 on RHS.

In next sections let’s jump into its definition, standard form, roots and methods to solve it.

## What is quadratic equation and examples

A quadratic equation is an expression in which there exists a quadratic polynomial on LHS with the highest
degree of 2 and 0 on RHS or in other words, if a quadratic polynomial equates to 0, then it becomes a quadratic
equation.

A quadratic equation in variable x is written in the form of ax^{2} + bx + c = 0, where a ≠ 0 and
a, b, c are real numbers.

2x^{2} + 4x + 1 = 0 is a quadratic equation

Because, on LHS, 2x^{2} + 4x + 1 is a quadratic polynomial with its highest degree 2 and
on RHS is 0 with sign of equality “=” between them.

4x^{2} -7x + 9 = 0 is an example of quadratic equation.

9 – x – 5x^{2} = 0 is also a quadratic equation.

## Standard form of quadratic equation

An equation whose terms are arranged in descending order of their degrees is known as standard form of
an equation.

The **standard form of quadratic equation** is
**ax ^{2} + bx + c = 0** where a, b and c are real numbers and a ≠ 0. The constants a, b and c are called as coefficients of the equation.

More specifically, a is called as quadratic coefficient, b is a linear coefficient and c is constant coefficient.

## Roots of quadratic equation

In a quadratic equation, the value of LHS becomes equal to RHS when it is 0. LHS can be set to 0 for some
values of x. When such numbers are put in the place of x to make the value of LHS 0, these specific values of
x are called as roots of the quadratic equation.

Such values of x are said to be satisfying the quadratic equation or “the values of x satisfy the
quadratic equation”.

For a quadratic equation, there always exist at most two roots.

## Methods to solve quadratic equation

Solving the quadratic equation means finding the roots of the quadratic equation. As explained above, roots
are the values of x which satisfy the equation ax^{2} + bx + c = 0. There are three ways to find
the roots or to solve the quadratic equation.

- Factorization
- Completing the square
- Quadratic formula

### 1. Factorization method

In this method, the middle term of the equation ax^{2} + bx + c = 0 split into two terms in such a
way that the sum of the two terms equals to the coefficient b and product of the two terms equals to the
coefficient c.

Here, the middle term is bx and b is the coefficient of bx. So, split the coefficient b into
two terms p and q such that p + q = b and p × q = c, where p ≠ 0 and q ≠ 0.

Therefore, after splitting b the equation ax^{2} + bx + c = 0 can be written as:

ax^{2 } + px + qx + c = 0

Now, LHS of ax^{2 } + px + qx + c = 0 can be expressed as product of two linear
factors. Each of the two linear factors can be equated to 0 separately.

After, solving each linear factor separately, we get one value of x from each of the two linear factors.

These two values of x will be the roots of quadratic equation.

Let’s understand it more clearly from the following examples.

**x² + 5x + 6 = 0**

Here a = 1, b = 5, c = 6

So, 5 can be split into 2 and 3

Because, 2 + 3 = 5, which is equal to value of b

and 2 × 3 = 6, which is equal to value of c

So, the middle term 5x can be split as 2x + 3x

Hence, x² + 5x + 6 = 0 can be written as:

x² + 2x + 3x + 6 = 0

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

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

So, here (x + 2) and (x + 3) are the two linear factors

(x + 2) and (x + 3) can be solved separately by equating them to zero

(x + 2) = 0
x = -2 |
(x + 3) = 0
x = -3 |

∴ -2 and -3 are the roots of x² + 5x + 6 = 0

### 2. Completing the square method

Following are the steps to solve the quadratic equation ax^{2} + bx + c = 0, a ≠ 0 using competing
the square method.
**Step 1: Make the coefficient of x ^{2} equal to 1 by dividing the whole equation by ‘a’**

$\frac{\mathrm{ax2}}{\mathrm{a}}$ + $\frac{\mathrm{bx}}{\mathrm{a}}$ + $\frac{\mathrm{c}}{\mathrm{a}}$ = 0

or x

^{2}+ $\frac{\mathrm{bx}}{\mathrm{a}}$ + $\frac{\mathrm{c}}{\mathrm{a}}$ = 0

**Step 2: Move the constant term**
$\frac{\mathrm{c}}{\mathrm{a}}$
on the other side.

x^{2} + +
$\frac{\mathrm{bx}}{\mathrm{a}}$
= 0 –
$\frac{\mathrm{c}}{\mathrm{a}}$

x^{2} + +
$\frac{\mathrm{bx}}{\mathrm{a}}$
= –
$\frac{\mathrm{c}}{\mathrm{a}}$

**Step 3: Half the coefficient of x i.e.**
$\frac{\mathrm{b}}{\mathrm{a}}$

$\frac{1}{2}\text{(}\frac{\mathrm{b}}{\mathrm{a}}\text{)}$
=
$\frac{\mathrm{b}}{\mathrm{2a}}$

**Step 4: Add the square of**
$\frac{\mathrm{b}}{\mathrm{2a}}$
on both sides of the equation

x^{2}
+
$\frac{\mathrm{bx}}{\mathrm{a}}$ +
${\text{(}\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^{2}$
=
$\u2013\frac{\mathrm{c}}{\mathrm{a}}$
+
${\text{(}\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^{2}$

**Step 5: Write LHS as perfect square of binomial expression and simplify RHS also**

${\text{(}\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^{2}$
=
$\frac{{\text{b}}^{2}\u2013\text{4ac}}{4{\text{a}}^{2}}$

**Step 6: Take square root on both sides**

$\sqrt{{\text{(}\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}\text{)}}^{2}}$
=
$\sqrt{\frac{{\text{b}}^{2}\u2013\text{4ac}}{4{\text{a}}^{2}}}$

**Step 7: On LHS, square root cancels out with the square. On RHS add ±**

$\text{x}+\frac{\mathrm{b}}{\mathrm{2a}}$
=
$\text{\xb1}\sqrt{\frac{{\text{b}}^{2}\u2013\text{4ac}}{4{\text{a}}^{2}}}$

**Step 8: Shift the constant term**
$\frac{\mathrm{b}}{\mathrm{2a}}$
from LHS to RHS

x
=
$\text{\xb1}\sqrt{\frac{{\text{b}}^{2}\u2013\text{4ac}}{4{\text{a}}^{2}}}$
–
$\frac{\mathrm{b}}{\mathrm{2a}}$

x
=
$\frac{\text{\xb1}\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$
–
$\frac{\mathrm{b}}{\mathrm{2a}}$

x
=
$\frac{\u2013\text{b}\text{\xb1}\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$

± in front of
$\text{\xb1}\sqrt{{\text{b}}^{2}\u2013\text{4ac}}$ indicates two values with + and – separately i.e.
$\sqrt{{\text{b}}^{2}\u2013\text{4ac}}$
and
$\u2013\sqrt{{\text{b}}^{2}\u2013\text{4ac}}$

So x has two values which can be written as
$\frac{\u2013\text{b}+\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$
and
$\frac{\u2013\text{b}\u2013\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$

These two values of x are called as two roots of the quadratic equation.

Roots of ax^{2} + bx + c = 0, a ≠ 0 are $\frac{\u2013\text{b}+\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$
and
$\frac{\u2013\text{b}\u2013\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$

**x² + 5x + 6 = 0**

Here, the coefficient of x² is 1

So, there is no need to divide the equation by 1 because dividing by 1 will have no effect
**x² + 5x = -6**

Now, add square of half of coefficient of x on both sides i.e.
${\text{(}\frac{5}{2}\text{)}}^{2}$

x² + 5x +
${\text{(}\frac{5}{2}\text{)}}^{2}$
= -6 +
${\text{(}\frac{5}{2}\text{)}}^{2}$

${\text{(}\text{x}+\frac{5}{2}\text{)}}^{2}$
=
-6 +
$\frac{25}{4}$

${\text{(}\text{x}+\frac{5}{2}\text{)}}^{2}$
=
$\frac{\mathrm{-24\; +\; 25}}{4}$

${\text{(}\text{x}+\frac{5}{2}\text{)}}^{2}$
=
$\frac{1}{4}$

$\text{x}+\frac{5}{2}$
=
$\pm \frac{1}{2}$

Solve for the value of x separately

$\text{x}+\frac{5}{2}$
=
$\frac{1}{2}$
$\text{x}=\frac{1}{2}\u2013\frac{5}{2}$ $\text{x}=\frac{\mathrm{1\; \u2013\; 5}}{2}$ $\text{x}=\frac{-4}{2}$ x = -2 |
$\text{x}+\frac{5}{2}$
=
$\u2013\frac{1}{2}$
$\text{x}=\u2013\frac{1}{2}\u2013\frac{5}{2}$ $\text{x}=\frac{\mathrm{-1\; \u2013\; 5}}{2}$ $\text{x}=\frac{-6}{2}$ x = -3 |

∴ x = -2 and x = -3 are the two roots of the equation.

### 3. Quadratic formula method

Quadratic formula method is another way to solve a quadratic equation. It is found easy
to use as compared to the factorization method and completing the square method.

The roots of quadratic equation a^{2} + bx + c = 0 are calculated using these two formulas

$\frac{\u2013\text{b}+\sqrt{\text{D}}}{\mathrm{2a}}$
and
$\frac{\u2013\text{b}\u2013\sqrt{\text{D}}}{\mathrm{2a}}$

Here, D is called **discriminant** and value of D =
${\text{b}}^{2}\u2013\text{4ac}$

The values of a, b and c are filled up in the above formulas to find roots. The values obtained after
solving it, will be the two roots of the quadratic equation.

The solution or roots of a quadratic equation remains always same when solved by any of the three methods Factorization method, Completing the square method and Quadratic formula method.

Let’s see next, how the quadratic formula method helps to solve a quadratic equation.

**x² + 5x + 6 = 0**

Here a = 1, b = 5, c = 6

Find discriminant, D

D =
${\text{b}}^{2}\u2013\text{4ac}$

=
${\text{(5)}}^{2}\u2013\text{4(1)(6)}$

= 25 – 24

= 1

Since D ≥ 0

Find the two roots using formula

∴ x =
$\frac{\u2013\text{b}+\sqrt{\text{D}}}{\mathrm{2a}}$
and
$\frac{\u2013\text{b}\u2013\sqrt{\text{D}}}{\mathrm{2a}}$

x =
$\frac{\u2013\text{5}+\sqrt{\text{1}}}{\mathrm{2(1)}}$
x = $\frac{\u2013\text{5}+1}{2}$ x = $\frac{-4}{2}$ x = -2 |
x =
$\frac{\u2013\text{5}\u2013\sqrt{\text{1}}}{\mathrm{2(1)}}$
x = $\frac{\u2013\text{5}\u20131}{2}$ x = $\frac{\mathrm{-6/mn>2}}{}$ x = -3 |

∴ -2 and -3 are the roots of the quadratic equation x² + 5x + 6 = 0

This method uses
Sridharacharay’s formula,
which gave the following two formulas to find the roots of the equation:

$\frac{\u2013\text{b}+\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$
and
$\frac{\u2013\text{b}\u2013\sqrt{{\text{b}}^{2}\u2013\text{4ac}}}{\mathrm{2a}}$

It was given by Indian mathematician
Sridharacharaya around 1025 AD.