The** quadratic formula** is used to find the roots of a quadratic equation. This formula helps to evaluate the solution of quadratic equations replacing the factorization method. If a quadratic equation does not contain real roots, then the quadratic formula helps to find the imaginary roots of that equation. The quadratic formula is also known as Shreedhara Acharya’s formula. In this article, you will learn the quadratic formula, derivation and proof of the quadratic formula, along with a video lesson and solved examples.

Let’s learn what a quadratic equation is and how to solve the quadratic equation using the quadratic formula.

In quadratic formula, b^{2} – 4ac is called the discriminant and is denoted by D.

## What is Quadratic Formula?

An algebraic expression of degree 2 is called the **quadratic equation**. The general form of a quadratic equation is ax^{2} + bx + c = 0, where a, b and c are real numbers, also called “**numeric coefficients” **and a ≠ 0. Here, x is an unknown variable for which we need to find the solution. We know that the **quadratic formula** used to find the solutions (or roots) of the quadratic equation ax^{2} + bx + c = 0 is given by:

## What is the Quadratic Formula used for?

The quadratic formula is used to find the roots of a quadratic equation and these roots are called the solutions of the quadratic equation. However, there are several methods of solving quadratic equations such as factoring, completing the square, graphing, etc.

## Roots of Quadratic Equation by Quadratic Formula

We know that a second-degree polynomial will have at most two zeros, and therefore a quadratic equation will have at most two roots.

In general, if α is a root of the quadratic equation ax^{2} + bx + c = 0, a ≠ 0; then, aα^{2} + bα + c = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation, ax^{2} + bx + c = 0.

**Note: **Roots of the quadratic equation ax^{2} + bx + c = 0 are the same as zeros of the polynomial ax^{2} + bx + c.

One of the easiest ways to find the roots of a quadratic equation is to apply the quadratic formula.

**Derivation of Quadratic Formula**

We can derive the quadratic formula in different ways using various techniques.

**Derivation Using Completing the Square Technique**

Let us write the standard form of a quadratic equation.

ax^{2} + bx + c = 0

Divide the equation by the coefficient of x^{2}, i.e., a.

x^{2} + (b/a)x + (c/a) = 0

Subtract c/a from both sides of this equation.

x^{2} + (b/a)x = -c/a

Now, apply the method of completing the square.

Add a constant to both sides of the equation to make the LHS of the equation as complete square.

Adding (b/2a)^{2} on both sides,

x^{2} + (b/a)x + (b/2a)^{2} = (-c/a) + (b/2a)^{2}

Using the identity a^{2} + 2ab + b^{2} = (a + b)^{2},

[x + (b/2a)]^{2} = (-c/a) + (b^{2}/4a^{2})

[x + (b/2a)]^{2} = (b^{2} – 4ac)/4a^{2}

Take the square root on both sides,

*x+b2a=±b2−4ac2a*

Therefore,

*x=−b±b2−4ac2a*

This is the most commonly used method to derive the quadratic formula in maths.

**Shortcut Method of Derivation**

Write the standard form of a quadratic equation.

ax^{2} + bx + c = 0

Multiply both sides of the equation by 4a.

4a(ax^{2} + bx + c) = 4a(0)

4a^{2}x^{2} + 4abx + 4ac = 0

4a^{2}x^{2} +4abx = -4ac

Add a constant on sides such that LHS will become a complete square.

Adding b^{2} on both sides,

4a^{2}x^{2} + 4abx + b^{2} = b^{2} – 4ac

(2ax)^{2} + 2(2ax)(b) + b^{2} = b^{2} – 4ac

Using algebraic identity a^{2} + 2ab + b^{2} = (a + b)^{2},

(2ax + b)^{2} = b^{2} – 4ac

Taking square root on both sides,

2ax + b = ±√(b^{2} – 4ac)

2ax = -b ±√(b^{2} – 4ac)

x = [-b ±√(b^{2} – 4ac)]/2a

How to Solve Using Quadratic Formula – Steps

Let us understand how to use the quadratic formula with the help of the steps given below. These steps will help you to understand the method of solving quadratic equations using the quadratic formula

Let us consider a quadratic equation:

**Step1: **Consider a quadratic equation x^{2} + 4x – 13 = 0

**Step 2: **Compare with the standard form and write the coefficients.

Here, a = 1, b = 4, c = -13

**Step 3:** Now. find the value of b^{2} – 4ac.

b^{2} – 4ac = (4)^{2} – 4(1)(-13) = 16 + 52 = 68

**Step 4:** Substitute the values in the quadratic formula to get the roots of the given quadratic equation.

x = [-b ± √(b^{2} – 4ac)]/ 2a

= [-4 ± √68]/ 2(1)

**Step 5: **Simplify the expression to get the values for x.

= [-4 ± 2√17]/2

= -2 ± √17

Therefore, x = -2 – √7 and x = -2 + √17 are the roots of the given quadratic equation.

Some of the important points about quadratic formula and the nature of roots of a quadratic equation are listed below:

- The expression under the radical in the quadratic formula is called the discriminant, i.e., D = b
^{2}– 4ac - The nature of the roots of a quadratic equation can be determined based on the value of D.

If D = 0, the two roots are real and equal

If D > 0, the roots are real and unequal

If D < 0, the roots are not real, i.e. imaginary - If the value of discriminant is 0, then the roots of the quadratic equation ax
^{2}+ bx + c = 0 are -b/2a and -b/2a.

Solved Examples on Quadratic Formula

**Example 1: Find the roots of the equation x**^{2 }**– 5x + 6 = 0 using the quadratic formula.**

^{2 }

**Solution:**

Given quadratic equation is:

x^{2 }– 5x + 6 = 0

Comparing the equation with ax^{2}+bx+c = 0 gives,

a = 1, b = -5 and c = 6

b^{2 }– 4ac = (-5)^{2 }– 4 × 1 × 6 = 25 – 24 = 1 > 0

The roots of the given equation are real.

Using quadratic formula,

x = [-b ± √(b^{2} – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. x = (5 + 1)/2 and x = (5 – 1)/2

x = 6/2, x = 4/2

x = 3, 2

Hence, the roots of the given quadratic equation are 3 and 2.

**Example 2: Find the roots of 4x**^{2}** + 3x + 5 = 0 using quadratic formula.**

^{2}

**Solution:**

Given quadratic equation is:

4x^{2} + 3x + 5 = 0

Comparing with the standard form ax^{2} + bx + c = 0,

a = 4, b = 3, c = 5

Determinant (D) = b^{2} – 4ac

= (3)^{2} – 4(4)(5)

= 9 – 80

= -71 < 0

That means, the roots are complex (not real).

Using quadratic formula,

x = [-b ± √(b^{2} – 4ac)]/ 2a

= [-3 ± √(-71)]/ 2(4)

= [-3 ± √(i2 71)]/ 8

= (-3 ± i√71)/8

Therefore, the complex roots of the given equation are x = (-3 + i√71)/8 and x (-3 – i√71)/8.

Quadratic Formula Questions

Solve with a quadratic formula and write the roots of equations.

- 2x
^{2}– 3x + 4 = 0

- x
^{2}+ 6x – 5 = 0

- 7x
^{2}+ 6x – 4 = 0