Solved Example on Quadratic Formula

Example 1: Write the quadratic function f(x) = (x – 9)(x + 3) in the general form of ax² + bx + c.

Solution: 

Given, the function as (x – 9)(x + 3)

= x² + 3x – 9x – 27

= x² – 6x – 27

This is the required form.

Example 2: Find the constants a, b, and c in the general form of equation 4x² + 5x + 9 = 0.

Solution: 

Given, equation is 4x² + 5x + 9 = 0….(1)

General form of quadratic equation is ax² + bx + c = 0.

Comparing the given equation with general equation we get,

a = 4, b = 5, c = 9.

Example 3: Write the quadratic function f(x) = (x + 8)(x – 3) in the general form of ax² + bx + c.

Solution:

Given, the function as (x + 8)(x – 3) 

= x² – 3x + 8x – 24

= x² + 5x – 24 [Which is the general form]

Example 4: Find the roots of equation 2x² – 4x + 2 = 0.

Solution: 

Here a = 2, b = -4, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.

⇒ Discriminant = b² – 4ac = (-4)² – 4(2)(2) 

⇒ Discriminant = 16 – 16 = 0 [which is equal to zero] 

So, it has real and equal roots.

Roots = (−b ± √(b² − 4ac)) / 2a

⇒ Roots = {-(-4) ± √(0) } / 2(2)

⇒ Roots = 4 / 4 = 1

Thus, 1 is the root of the equation.

Example 5: Find the roots of equation 4x² – 3x + 3.

Solution: 

Here a = 4, b = -3, c = 3, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.

Discriminant = b² – 4ac 

⇒ Discriminant = (-3)² – 4(4)(3) 

⇒ Discriminant = 9 – 48 = -39  [which is negative]

So, it has complex roots.

Roots = {−b ± √(b² − 4ac)] / 2a

⇒ Roots = (-(-3) ± √-39 / 2(4) )

⇒ Roots = (3 ± 39i)/8 

Thus, (3 + 39i)/8 and (3 – 39i)/8 are the roots of the quadratic equation.

Example 6: Find the roots of the quadratic equation 6x² – 8x + 2 = 0.

Solution:

Here a = 6, b = -8, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.

Discriminant = b² – 4ac 

⇒ Discriminant = (-8)(-8) – 4(6)(2) 

⇒ Discriminant = 64 – 48 = 16 [which is positive] 

So, it has real and distinct roots.

x = (-b ± √ (b² – 4ac) )/2a

⇒ x = (-(-8) ± √(16) / 2(6)

⇒ x = (8 ± √16) / 12

⇒ x =  (8 ± 4)12

Taking +ve sign, we get x = 1, and 

Taking -ve sign, we get x = 1/3.

Thus, 1/3 and 1 are the roots of the equation.

Quadratic Formula is used to find the roots (solutions) of any quadratic equation. Using the Quadratic formula real and imaginary all the types of roots of the quadratic equations are found.

The quadratic formula was formulated by a famous Indian mathematician Shreedhara Acharya, hence it is also called Shreedhara Acharya’s Formula. It is used to find the solution of the quadratic equation of the form ax² + bx + c = 0. So, let’s start learning about the concept of Quadratic Formula.