Quadratic Formula

The quadratic formula is used to find the roots (solutions) of any quadratic equation. It helps in finding all types of roots, including real and imaginary roots.

• 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, where a ≠ 0, and a, b, and c are real constants.

Example: Find the roots of the equation x^2 + 5x + 6 = 0.

For x^2 + 5x + 6 = 0, use the quadratic formula: a = 1, b = 5, and c = 6.

Substitute into the formula:

x = \frac{-5\pm\sqrt{5^2-4(1)(6)}}{2(1)}

x = \frac{-5\pm\sqrt{25-24}}{2}

x = \frac{-5\pm 1}{2}

The two solutions are: x = −2 and x = −3

Discriminant of a Quadratic Equation

The discriminant of a quadratic equation ax² + bx + c = 0 is denoted by D and given by,

D = b² - 4ac

Discriminant of a Quadratic Equation is very helpful in determining the nature of the root of quadratic equations.

Nature of Root of Quadratic Equation

To find the nature of the roots of the quadratic equation we find the discriminant of the given quadratic equation. The term is called discriminant because it determines the nature of the roots of the quadratic equation based on its sign.

There are 3 types in the nature of roots, 

Maximum and Minimum Value of Quadratic Expression

The maximum and minimum values for the quadratic equation of the form ax² + bx + c = 0 can be observed with the help of graphs.

• If the value of a is positive i.e. (a > 0), the quadratic equation has a minimum value at x = -b/2a i.e., -D/4a.
• If the value of a is negative i.e. (a < 0), the quadratic equation has a maximum value at x = -b/2a i.e., -D/4a.

where D is the discriminant of the Quadratic Expression.

Derivation of the Quadratic Formula

We can easily derive the quadratic formula using two methods 

1. By Completing the Square Technique

Derivation of the quadratic formula is achieved by using the Completing Square method

Let us take the standard form of a quadratic equation i.e. ax² + bx + c = 0

Dividing the equation by the coefficient of x², i.e., x² + (b/a)x + (c/a) = 0

Subtracting c/a from both sides: x² + (b/a)x = -c/a

Now, by completing the square method,

We have to add a specific constant to both sides of the equation to make the LHS a complete square.

Here, we add (b/2a)² to both sides of the equation

x² + (b/a)x + (b/2a)² = (-c/a) + (b/2a)²

Using a² + 2ab + b² = (a + b)²,

⇒ [x + (b/2a)]² = (-c/a) + (b²/4a²)
⇒ [x + (b/2a)]² = (b² – 4ac)/4a²

Taking square root on both sides, [x + (b/2a)] = √[(b² – 4ac)] / 2a

Now, subtracting b/2a from both sides we get

x=\frac{-b\pm \sqrt{b^2-4*a*c}}{2a}

This is the required quadratic formula.

2. Shortcut Method of Derivation

The quadratic formula is derived by the shortcut method as,

The given standard form of a quadratic equation is,

ax² + bx + c = 0

Multiply both sides by 4a

4a(ax² + bx + c) = 4a × (0)

⇒ 4a²x² + 4abx + 4ac = 0

⇒ 4a²x² +4abx = -4ac

By completing the square method, add b² on both sides.

4a²x² + 4abx + b² = b² – 4ac

⇒ (2ax)² + 2(2ax)(b) + b² = b² – 4ac    {We know that, a² + 2ab + b² = (a + b)²)}

⇒ (2ax + b)² = b² – 4ac

Taking square root

2ax + b = ±√(b² – 4ac)

⇒ 2ax = -b ±√(b² – 4ac)

x=\frac{-b\pm \sqrt{b^2-4*a*c}}{2a}

This proves the quadratic formula.

Roots of Quadratic Equation by Quadratic Formula

A quadratic equation is a two-degree quadratic equation and it has two a maximum of two roots. They can either be real, or imaginary. 

The roots of the quadratic equation is the value that satisfies the quadratic equation thus, we can say that if the given quadratic equation is, ax² + bx + c = 0, and if α is a root of the quadratic equation, then α satisfies the quadratic equation, i.e., aα² + bα + c = 0. We also define roots of the quadratic equations ax² + bx + c =  as the zeros of the polynomial ax² + bx + c.

We can easily find the roots of the quadratic equation by using the following quadratic formula,

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

This gives the two roots of the quadratic equation, 

Taking positive(+)Sign

x=\frac{-b\pm \sqrt{b^2-4*a*c}}{2a}

Taking negative(-)Sign

xx=\frac{-b\pm \sqrt{b^2-4*a*c}}{2a}

Thus using the quadratic formula we easily get the roots of the quadratic equation.

Example: Simplify x² + x - 6

Step 1: Comparing x² + x - 6 with ax²+ bx + c = 0

Step 2: a = 1, b = 1, and c = -6

Step 3: Using the quadratic formula,

x=\frac{-b\pm \sqrt{b^2-4*a*c}}{2a}

Step 4: Simplifying

x = [-b ± √(b² – 4ac)] / 2a
⇒ x = [-1 ± √(1² – 4(1)(-6))] / 2(1)
⇒ x = [-1 ± √(4 + 4(1)(6))] / 2(1)
⇒ x = (-1 ± 5)/2
⇒ x = (-1-5)/2 = -3
⇒ x = (-1+5)/2 = 2

Thus, the values of x are x = -3 and x = 2

Also Check

• Quadratic Function
• Quadratic Inequalities
• Solving Cubic Equations

Solved Example on Quadratic Formula

Example 1: Write the quadratic function f(x) = (x - 9)(x + 3)in the general form ofax² + bx + c.

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.

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.

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.

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.

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.

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.