Quadratic Equation

Quadratic equation is the second-degree equation in which one variable contains the variable with an exponent of 2.

Its general form is ax2 + bx + c = 0, a ≠ 0

Types of Quadratic Equations

There are following two special types of quadratic equations: –

  1. Quadratic Equations with no term in x
  2. Quadratic Equations with no constant term

See also: Types of Equations

1. Quadratic Equations with no term in x:

If there is a quadratic equation with no term in x then we can move the constant to the other side. For example,

Example:

Solve x2 – 2 = 0

Solution:

x2 – 2 = 0

x2 = 2

x = 2 or x = –2

2. Quadratic Equations with no constant term:

Example:

Solve x2 – 16x = 0

Solution:

x2 – 16x = 0

By taking common x

x(x – 16) = 0

As the product of the two factors is 0, therefore, one or both of the factors is zero

x(x) – 16 = 0. Hence, x = 0 or x – 16 = 0

Therefore, the two solutions are

x = 0, or x = 16

You may also read: Linear Equation

Methods of solving Quadratic Equations:

Quadratic equations can be solved by using following four methods:

  1. By factorization
  2. By completing the square
  3. By using quadratic formula
  4. By using graphs

But, here we discussed the first three methods of solving the quadratic equations. However, solution of quadratic equation by factorization and use of quadratic formula are particularly significant.

a) Solution of Quadratic Equations by Factorization

The procedure of solving any Quadratic Equation by factorization is same as used in above example that if we find 0 as the product of two numbers then at least one of the numbers must be 0.

This means that, if AB = 0 then A = 0 or B = 0

In Factorization, first we learn, how to factorize monic quadratic equation then we learn how to factorize a non-monic quadratic equation.

A non-monic quadratic equation is an equation of the form x2 + bx + c. we always try to find out two numbers whose sum is ‘b’ and whose product is ‘c’. Now we apply this method to solve some monic quadratic equations.

Example No.1:

Solve x2 – 6x – 16 = 0

Solution:

First, we try to find out two numbers whose product is 16 and whose sum is –6. obviously, –8, + 2 are our required numbers. So, we can factor as:

x2 – 6x + 16 = 0

x2 – 8x + 2x – 16 = 0

x(x – 8) + 2(x – 8) = 0

(x – 8)(x + 2) = 0

As the product of the two factors is 0, therefore, one or both of the factors is zero

x – 8 = 0 or x + 2 = 0

Therefore,

x = 8 or x = 2

Example No.2:

Solve x2 – 7x + 12 = 0

Solution:

First, we try to find out two numbers whose product is 12 and whose sum is –7. obviously, –4, –3 are our required numbers. So, we can factor as:

x2 – 7x + 12 = 0

x2 – 4x – 3x +12 = 0

x(x – 4) – 3(x – 4) = 0

(x – 4)(x – 3) = 0

As the product of the two factors is 0, therefore, one or both of the factors is zero

x – 4 = 0 or x – 3 = 0

Therefore,

x = 4 or x = 3

A non-monic quadratic equation is an equation of the form ax2 + bx + c. we utilize the same method to solve a non-monic quadratic equation. Now we apply this method to solve few non-monic quadratic equations.

Example No.1:

Solve 2x2 – 5x – 12 = 0

Solution:

Here, we multiple 2 and 12 to give 24 and find the two numbers to give 24 and add to give –5 numbers. Obviously, –8, +3 are our required numbers. So, we can factor as:

2x2 – 5x – 12 = 0

2x2 – 8x + 3x – 12 = 0

2x(x – 4) + 3(x – 4) = 0

(x – 4)(2x + 3) = 0

As the product of the two factors is 0, therefore, one or both of the factors is zero

x – 4 = 0 or 2x + 3 = 0

Therefore,

x = 4 or x = –3/2

Example No.2:

Solve 3x2 + 25x – 18 = 0

Solution:

Here, we multiple 3 and 18 to give 54 and find the two numbers to give 54 and add to give +25 numbers. Obviously, +27, –2 are our required numbers. So, we can factor as:

3x2 + 25x – 18 = 0

3x2 + 27x – 2x – 18 = 0

3x(x + 9) – 2(x + 9) = 0

(x + 9)(3x – 2) = 0

As the product of the two factors is 0, therefore, one or both of the factors is zero

x + 9 = 0 or 3x – 2 = 0

Therefore,

x = –9 or x = 2/3

b) Solution of Quadratic Equations by completing the square

Now we are familiar with quadratic equations having one or two solutions that were rational but there are several quadratic questions that have irrational solutions or in few cases, no real solutions. Therefore, in order to handle the more general quadratic questions, we employ a most commonly method of solution, which is known as completing the square. This method can also be utilized for finding the radius and centre of a circle, minimum or maximum of a quadratic function, putting integrals into typical form in Calculus, etc.

Example No.1:

Solve x2 + 2x − 10 = 0

Solution:

            First, we focus on x2 + 2x to find a number that must be added to x2 + 2x to make the expression into a perfect square. Take half of the coefficient and square it in both cases i.e. x positive or x negative, here it is 1, therefore, x2 + 2x + 1 = (x + 1)2. Hence,

x2 + 2x − 10 = (x2 + 2x + 1) – 1 – 10 (add and subtract 1)

= (x + 1)2 – 11

(x + 1)2 = 11

Now take the positive and negative square roots to obtain

x + 1 = √(11 ) or x + 1 = – √11

x = – 1 + √(11 ) or x= – 1 – √(11 )

Example No.2:

Solve x2 + 4x − 10 = 0

Solution:

x2 + 4x − 10 = 0

x2 + 4x + 4 – 4 − 10 = 0   (complete the square by utilizing the technique mentioned in above example )

(x + 4)2 = 4 + 10

(x + 4)2 = 14

x + 4 = 14 or x + 4 = −14

Therefore,

x = 10 or x = −18

c) Solution of Quadratic Equations by using Quadratic Formula

General quadratic equation is ax2 + bx + c = 0

We utilize this formula to solve the given equation,

X = (-b±√(b^2-4ac)) / 2a

Now, we apply this formula to solve few quadratic equations.

Example No.1:

Solve x2 − 9x − 4 = 0

Solution:

x2 − 9x − 4 = 0

Here, a = 1, b = -9 and c = -4

Therefore,

X = (-9±√(81-4(-4))) ⁄ (2(1))

X = (-9±√(81-16)) / 2

X = (-9±√65) / 2