** 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

**ax**

^{2}**. So, let’s start learning about the concept of Quadratic Formula.**

**+ bx + c = 0**Table of Content

## What is a Quadratic Function?

A quadratic function is a polynomial function with one or more variables in which the highest exponent of the variable is two.

The standard form of the quadratic equation is **f(x) = ax**^{2 }** + bx + c**, this says that at least one term in the given equation is squared. In the above equation

**are constant terms and**

**a, b, and c****is a variable.Â**

**x**

f(x) = ax^{2}+ bx + cWhere a is not equal to 0 and a, b, and c are real numbers.

Some examples of quadratic expressions are:

- 4x
^{2}+ 3x + 5 - 6x
^{2}+ x + 7 - 7x
^{2}+ 5x

** Note: **Since the highest degree term in a quadratic function is of the second degree, therefore it is also called the polynomial of degree 2.

### Methods to Solve Quadratic Equations

A quadratic equation is solved to find the two values of x which satisfies the equation these values are also called the two roots of the quadratic equation.

There are various methods to find the roots of a quadratic equation. The important methods are:

- Quadratic Formula Method
- Factorization Method
- Completing the Square Method
- Graphical Method

In this article, we will only study the Quadratic (Shreedhara Acharya’s) Formula method.

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## What is Quadratic Formula?

For equation, f(x) = ax^{2}+ bx + c where a is not equal to 0 and a, b, and c are real numbers; solution of f(x) = 0 is given by Quadratic Formula i.e.,

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

The term under the square root i.e., b^{2} – 4ac, is called the discriminant, which determines the nature of the roots:

- discriminant > 0, the equation has two real and distinct roots.
- discriminant = 0, the equation has exactly one real root (a repeated root).
- discriminant < 0, the equation has two complex (or imaginary) roots.

## Derivation of Quadratic Formula

We can easily derive the quadratic formula, using two methodsÂ

- By Completing the Square Technique
- Shortcut Method of Derivation

Now let’s learn about them in detail.

### 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.

ax^{2} + bx + c = 0

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

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

Subtracting c/a from both sides.

x^{2} + (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

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

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

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

â‡’ [x + (b/2a)]^{2} = (b^{2} â€“ 4ac)/4a^{2}

Taking square root on both sides,

[x + (b/2a)] = âˆš[(b^{2} â€“ 4ac)] / 2a

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

x = [-b Â± âˆš(b^{2}Â â€“ 4ac)] / 2a

This is the required quadratic formula.

### Shortcut Method of Derivation

The quadratic formula is derived by the shortcut method as,

The given standard form of a quadratic equation is,

ax^{2} + bx + c = 0

Multiply both sides by 4a

4a(ax^{2} + bx + c) = 4a Ã— (0)

â‡’ 4a^{2}x^{2} + 4abx + 4ac = 0

â‡’ 4a^{2}x^{2} +4abx = -4ac

By completing the square method, add b^{2} on both sides.

4a^{2}x^{2} + 4abx + b^{2} = b^{2} â€“ 4ac

â‡’ (2ax)^{2} + 2(2ax)(b) + b^{2} = b^{2} â€“ 4ac Â Â {We know that, a^{2} + 2ab + b^{2} = (a + b)^{2})}

â‡’ (2ax + b)^{2} = b^{2} â€“ 4ac

Taking square root

2ax + b = Â±âˆš(b^{2} â€“ 4ac)

â‡’ 2ax = -b Â±âˆš(b^{2} â€“ 4ac)

x = [-b Â±âˆš(b^{2}â€“ 4ac)]/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.Â

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, ax^{2} + bx + c = 0, and if Î± is a root of the quadratic equation, then Î± satisfies the quadratic equation, i.e., aÎ±^{2} + bÎ± + c = 0. We also define roots of the quadratic equations ax^{2} + bx + c = Â as the zeros of the polynomial ax^{2} + bx + c.

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

[Tex]\bold{x = \frac{-b \pm \sqrt{b^2 â€“ 4ac}}{2a}} [/Tex]

This gives the two roots of the quadratic equation,Â

**Taking +ve Sign**

x = [-b + âˆš(b^{2}Â â€“ 4ac)] / 2a

**Taking -ve Sign**

x = [-b – âˆš(b^{2}Â â€“ 4ac)] / 2a

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

### Discriminant of a Quadratic Equation

The discriminant of a quadratic equation ax^{2} + bx + c = 0 is denoted by D and given by,

D = b^{2}– 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

**because it determines the nature of the roots of the quadratic equation based on its sign.**

**discriminant**There are ** 3 types** in the nature of roots,Â

For real and distinct roots, the discriminant should be positive i.e.**Real and distinct roots:****b**^{2}The curve of the equation intersects the x-axis at two different points.**– 4ac > 0.**For real and equal roots, the discriminant is zero i.e.**Real and equal roots:****b**^{2}The curve of the equation intersects the x-axis at only one point.**– 4ac = 0.**For complex roots, the discriminant is negative i.e.**Complex roots:****b**^{2}The curve of the equation does not intersect the x-axis.**– 4ac = 0.**

### Maximum and Minimum Value of Quadratic Expression

The maximum and minimum values for the quadratic equation of the form ax^{2} + 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.

## What is Quadratic Formula used for?

The quadratic function formula is the most used formula in algebra it is used for solving quadratic equations and finding their roots instantly without performing various steps. It gives the solution of the quadratic equations where the normal solution is not possible. We use this formula for solving various types of problems in algebra, factoring equations, graphing various curves, and others.

### How to Solve Quadratic Equation Using Quadratic Formula?

Follow the steps given below to solve Using Quadratic Formula

Write the given equation in standard form as, axStep 1:^{2}+ bx + c = 0

Carefully note the coefficient from the above equation as, a, b and c.Step 2:

Use the Quadratic Formula, x = [-b Â± âˆš(bStep 3:^{2}Â â€“ 4ac)] / 2a

Put all the values of a, b and c and simplify for x.Step 4:

We can understand this with the help of the following example.

**Example: Simplify x**^{2}** + x – 6**

**Solution:**

Comparing xStep 1:^{2}+ x – 6 with ax^{2}+ bx + c = 0

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

Using the quadratic formula,Step 3:x = [-b Â± âˆš(b

^{2}Â â€“ 4ac)] / 2a

SimplifyingStep 4:x = [-b Â± âˆš(b

^{2}Â â€“ 4ac)] / 2aâ‡’ x = [-1 Â± âˆš(1

^{2}Â â€“ 4(1)(-6))] / 2(1)â‡’ x = [-1 Â± âˆš(4Â + 4(1)(6))] / 2(1)

â‡’ x = (-1 Â± 5)/2

â‡’ x = (-1-5)/2 = -3

â‡’ x = (-1+5)/2 = 2

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

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## Solved Example on Quadratic Formula

**Example 1: Write the quadratic function f(x) = (x – 9)(x + 3)****in the general form of****ax**^{2}** + bx + c.**

**Solution:Â **

Given, the function as (x – 9)(x + 3)

= x

^{2}+ 3x – 9x – 27= x

^{2}– 6x – 27This is the required form.

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

**Solution:Â **

Given, equation is 4x

^{2}+ 5x + 9 = 0….(1)General form of quadratic equation is ax

^{2}+ 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 ax**^{2}** + bx + c.**

** Solution**:

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

= x

^{2}– 3x + 8x – 24= x

^{2}+ 5x – 24 [Which is the general form]

**Example 4: Find the roots of equation 2x**^{2}** – 4x + 2 = 0.**

**Solution:Â **

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.

â‡’ Discriminant = b

^{2}– 4ac = (-4)^{2}– 4(2)(2)Ââ‡’ Discriminant = 16 – 16 = 0 [which is equal to zero]Â

So, it has real and equal roots.

Roots = (âˆ’b Â± âˆš(b

^{2 }âˆ’ 4ac)) / 2aâ‡’ Roots = {-(-4) Â± âˆš(0) } / 2(2)

â‡’ Roots = 4 / 4 = 1

Thus, 1 is the root of the equation.

**Example 5: Find the roots of equation 4x**^{2}** – 3x + 3.**

**Solution:Â **

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 = b

^{2}– 4acÂâ‡’ Discriminant = (-3)

^{2}– 4(4)(3)Ââ‡’ Discriminant = 9 – 48 = -39 Â [which is negative]

So, it has complex roots.

Roots = {âˆ’b Â± âˆš(b

^{2 }âˆ’ 4ac)] / 2aâ‡’ Roots = (-(-3) Â± âˆš-39 / 2(4) )

â‡’ Roots = (3 Â± 39i)/8Â

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

**Example 6: Find the roots of the quadratic equation 6x**^{2}** – 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 = b

^{2}– 4acÂâ‡’ Discriminant = (-8)(-8) – 4(6)(2)Â

â‡’ Discriminant = 64 – 48 = 16 [which is positive]Â

So, it has real and distinct roots.

x = (-b Â± âˆš (bÂ² – 4ac) )/2a

â‡’ x = (-(-8) Â± âˆš(16) / 2(6)

â‡’ x = (8 Â± âˆš16) / 12

â‡’ x = Â (8 Â± 4)12

Taking +ve sign, we get x = 1, andÂ

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

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

## FAQs on Quadratic Formula

**What is a Quadratic Equation?**

**What is a Quadratic Equation?**

Any second-degree equation of the form ax

^{2}+ bx + c = 0 is called the quadratic equation. Where a, and b, are the coefficients, c is the constant term, with x as the variable.Â

**What is the determinant in the Quadratic Formula?**

**What is the determinant in the Quadratic Formula?**

The value b

^{2}– 4ac is called the discriminant of the quadratic formula and is denoted by the symbol D.

**Who invented the quadratic formula?**

**Who invented the quadratic formula?**

The quadratic formula was invented by the famous Indian mathematician called Shridhara Acharya.

**What is the Quadratic Formula?**

**What is the Quadratic Formula?**

The Quadratic formula is used to solve quadratic equations it is given as,

x = [-b Â± âˆš(b^{2}Â â€“ 4ac)]/ 2a

**When to use the quadratic formula?**

**When to use the quadratic formula?**

The quadratic formula is used when we are unable to find the roots of the quadratic equations by normal methods. This formula is also used if the roots of the quadratic equation are imaginary.