A quadratic equation is a polynomial whose highest exponent is the square of a variable. Standard form of quadratic equation looks like ax2 + bx + c = 0, where a ≠ 0 and a, b, c are numbers. The quadratic equation PDF is also provided here which the students can download and refer offline. This is one of the most important topics in maths. Among the various concepts of quadratic equation, the key points that are covered here are:

• The standard form of Quadratic Equation
• The solution for Quadratic equation using factoring the quadratic equation and completing the square method
• Nature of Roots of a Quadratic Equation
• The relation between Coefficient and Roots of the Quadratic Equation

A quadratic equation is defined as any equation which has a degree of 2, also called as “equation of degree 2”. The word “Quadratic” comes from “quad” means four and “Quadratic” means ‘to make a square’.

## Standard Form of Quadratic equation:

The standard form of a polynomial equation having a degree of two is $ax^{2}+bx+c = 0$ $a \neq 0$ Where “x” is a variable and a, b, c are real numbers.

Note: If a = 0 then the equation will become a linear equation rather than a quadratic equation.

Sridhara’s Method, one of the most common methods for finding the roots.

$ax^{2}+bx+c = 0$, which is given as:

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

here $\pm$ represent both the values (one positive and one negative), hence there are two possible roots for the equation.

Depending upon the value of discriminant, the quadratic equation can have 3 different types of roots, namely:

 S.no Value of Discriminant (D) Nature of Roots 1 $b^{2}-4ac > 0$ Two distinct Real roots 2 $b^{2}-4ac = 0$ Two equal roots 3 $b^{2}-4ac < 0$ Two distinct Complex Roots

As a quadratic equation has degree of 2, hence the number of solution that will satisfy the given equation will also be equal to 2. There can be more than one method to solve quadratic equations:

• Sridhara’s Formula
• Harriot’s Method

 Example: Find the roots of the equation $2x^{2} + 7x + 3 = 0$ Solution: Given: $2x^{2} + 7x + 3 = 0$ Here a = 2, b = 7 and c = 3 Finding the value of D, $D = b^{2}-4ac = 7^{2}- 4\times 2 \times 3 = 49 – 24$$\Rightarrow D = 25$ As $D >0$, the equation has two distinct real roots. $x = \frac{-b \pm \sqrt{D}}{2a}$ $\Rightarrow x = \frac{-7 \pm \sqrt{25}}{2 \times 2}$ $\Rightarrow x = \frac{-7 \pm 5}{4}$ $\Rightarrow x = \frac{-7 – 5}{4}\; or \; \frac{-7 + 5}{4}$ $\Rightarrow x = -3 \; or \; \frac{-1}{2}$ Example 2: Find the roots of the equation: x2  + 2x – 3 = 0 Solution: Given equation is x2  + 2x – 3 = 0 Let’s use the  factorization method to find its roots x2 – x + 3x – 3 = 0 x(x – 1) + 3(x – 1) = 0 (x – 1)(x + 3) = 0 Therefore  x = 1 or x = -3 are the roots of the given equation.

Quadratic equations are used to represent many mathematical and physical problems. A parabola is one of the real examples of quadratic equations.

Quadratic equation helps to find the position of an object travels in the air and again hit the surface back, loss or profit for items sold, speed and distance covered by a boat in the water against upstream or downstream and many more.

Problem 1: Vicky agreed to paint a wall for Rs. 150. It took him two hours longer than he expected, and therefore he earned Rs. 2.50 per hour less than anticipated. How long did he expect the painting would take?

Solution: Let m be the number of hours he expected. Then m+2 the total number of hours actually took to complete the painting. As he earned Rs. 2.50 per hour less. The rate of pay is divided by the number of hours.

Using above details, we have equation

150/x – 5/2 = 150/(x+2)

(300 – 5x)/2x = 150/(x+2)

300(x+2) – 5x(x+2) = 300x

300 x + 600 – 5x2 – 10x = 300x

-5x2 – 10x – 600 = 0

x2 + 2x – 120 = 0

(x + 12)(x – 10) = 0

or x = -12 or x = 10

Ignore the negative value as not valid. Therefore he anticipated that the painting would take 10 hours.

Problem 2: Krishan bought a land whose length is 10 times the width. If the land is expanded in the 1000 square meter area. What would be the length and width of the land.

Solution: Let x be the width of the land. Then 10x be its length. Also we are given with area of the land which is 1000 m2.

Therefore, 10x2 = 1000

x2  = 100

x = 10 or  x = -10

Ignore negative value as length cannot be negative.

Therefore, the width is 10 m and length is 10(10) = 100 m.