**Quadratics** or **quadratic equations** can be defined as a polynomial equation of a second degree, which implies that it comprises of minimum one term that is squared. The definite form is **axÂ² + bx + c = 0;** where x is an unknown variable and a,b,c are numerical coefficients Here, a â‰ 0 because if it equals to zero then the equation will not remain quadratic anymore and it will become a linear equation, such as bx+c=0.

The terms a, b and c are also called quadratic coefficients. The solutions to the quadratic equation are the values of unknown variable x, which satisfy the equation. These solutions are called as **roots or zeros** of quadratic equations. It means that, if we put the value of roots in the given quadratics, L.H.S. will be equal to R.H.S. of the equation. The roots of any polynomial are the solutions for the given equation.

## Quadratic Equation Definition

The polynomial equation whose highest degree is two, is called a quadratic equation. It is expressed in the form of:

**axÂ² + bx + c = 0**

where x is the unknown variable and a, b and c are the constant terms.

Since the quadratic include only one unknown term or variable, thus it is called univariate. The power of variable x are always non-negative integers, hence the equation is a polynomial equation with highest power as 2.

The solution for this equation is the values of x, which are also called as zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x in the Left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.

**Also, read:**

### Quadratic Equation Formula

The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose,Â **axÂ² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:**

**x = [-bÂ±âˆš(b ^{2}-4ac)]/2**

The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.

### Quadratics Equation Examples

Beneath are the illustrations of quadratic equations of the form (axÂ² + bx + c = 0)

- xÂ² â€“x â€“ 9 = 0

- 5xÂ² â€“ 2x â€“ 6 = 0

- 3xÂ² + 4x + 8 = 0

- -xÂ² +6x + 12 = 0

Examples of a quadratic equation with the absence of a â€˜ C â€˜- a constant term.

- -xÂ² â€“ 9x = 0

- xÂ² + 2x = 0

- -6xÂ² â€“ 3x = 0

- -5xÂ² + x = 0

- -12xÂ² + 13x = 0

- 11xÂ² – 27x = 0

Following are the examples of a quadratic equation in factored form

- (x â€“ 6)(x + 1) = 0Â [ result obtained after solving is xÂ² â€“ 5x â€“ 6 = 0]

- â€“3(x â€“ 4)(2x + 3) = 0 Â [result obtained after solving is -6xÂ² + 15x + 36 = 0]

- (x âˆ’ 5)(x + 3) = 0 Â [result obtained after solving is xÂ² âˆ’ 2x âˆ’ 15 = 0]

- (x – 5)(x + 2) = 0 Â [ result obtained after solving is xÂ² – 3x – 10 = 0]

- (x – 4)(x + 2) = 0 Â [result obtained after solving is xÂ² – 2x – 8 = 0]

- (2x+3)(3x – 2) = 0 Â [result obtained after solving is 6xÂ² + 5x – 6]

Below are the examples of a quadratic equation with an absence of linear co – efficient â€˜ bxâ€™

- 2xÂ² â€“ 64 = 0

- xÂ² â€“ 16 = 0

- 9xÂ² + 49 = 0

- -2xÂ² â€“ 4 = 0

- 4xÂ² + 81 = 0

- -xÂ² â€“ 9 = 0

### Solution of Quadratics by Factorisation

- Begin with a equation of the form axÂ² + bx + c = 0
- Ensure that it is set to adequate zero.
- Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
- Assign each factor equal to zero.
- Now solve the equation in order to determine the values of x.

Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.

2xÂ²-x-6=0

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

2x+3=0

x=-3/2

x=2

Also, Sridhara wrote down rules for Solving Quadratic Equation, therefore the most common method of finding the roots of the quadratic equation is known as Sridharacharya rule.

For the given Quadratic equation of the form,Â **axÂ² + bx + c = 0**

Therefore the roots of the given equation can be found by:

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

where \(\pm\) (one plus and one minus) represent two distinct roots of the given equation.

Lets Work Out:
Solving the quadratic equation using the above method: \(x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) \(x = \frac{-(-5)\pm \sqrt{(-5)^{2} -4 \times 3 \times 2}}{2 \times 3}\) \(x = \frac{5 \pm 1}{6}\) \(x = \frac{6}{6} \;\; or \;\; \frac{4}{6}\) or, \(x = 1 \;\; or \;\; \frac{2}{3}\)< |

The video below explains the concept of quadratics in a personalised manner for students to understand the concept in an easy way.