We know that, in mathematics calculations and solving problems are made easier by using formulae. In the same way, quadratic equation can be easily solved by using a formula. **The quadratic formula, which is very useful for finding its roots can be derived using the method of completing the square. **Let us derive the quadratic formula and learn how to use it for finding roots of the quadratic equations.

## (c) Solution of a quadratic equation by formula method:

Consider the quadratic equation ax^{2} + bx + c = 0, a≠0

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

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

Find half the coefficient of x and square it,

^{1}/_{2} x ^{b}/_{a} = ^{b}/_{2a}

(^{b}/_{2a})^{2} = ^{b^2}/_{(4a)^2}

Transpose the constant ^{c}/_{a} to RHS

x^{2} + ^{b}/_{a} x = – ^{c}/_{a}

Add (^{b}/_{2a})^{2 }to both sides of the equation

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

Factorise LHS and simplify RHS

(x + ^{b}/_{2a})^{2} = ^{b^2}/_{4a^2} – ^{c}/_{2a}

Take square root on both sides of the equation

x + ^{b}/_{2a} = ± √(^{b^2 – 4ac}/_{4a^2}) = ±^{√(b^2 – 4ac)}/_{2a}

x = –^{b}/_{2a} ±^{√(b^2 – 4ac)}/_{2a}

x = ^{-b±√(b^2 – 4ac)}/_{2a}

Therefore, the roots of the quadratic equation ax^{2} + bx + x = 0 are ^{–b – √(b^2 – 4ac)}/_{2a} and ^{–b + √(b^2 – 4ac)}/_{2a}

x = ^{-b±√(b^2 – 4ac)}/_{2a} is known as quadratic formula.

**Example: Solve the quadratic equation x**^{2} – 7x + 12 = 0 by formula method

^{2}– 7x + 12 = 0 by formula method

Solution:

Given x^{2} – 7x + 12 = 0

The given quadratic equation x^{2} – 7x + 12 = 0 is of the form ax^{2} + bx + c = 0 where a = 1, b = – 7 and c = 12

## Quadratic Equation – Exercise 9.5 – Class X

### Solve the following quadratic equations by using the formula method.

- x
^{2}– 4x + 2 = 0 - x
^{2}– 2x + 4 = 0 - 2y
^{2}+ 6y = 3 - 15m
^{2}– 11m + 2 = 0 - 8r
^{2}= r + 2 - (2x + 3)(3x – 2) + 2 = 0
- a(x
^{2}+ 1) = x(a^{2}+ 1) - x
^{2}+ 8x + 6 = 0

### Quadratic Equation – Exercise 9.5 – Solution:

**Solve the following quadratic equations by using the formula method.**

**x**^{2}– 4x + 2 = 0

Solution:

x^{2} – 4x + 2 = 0

**x**^{2}– 2x + 4 = 0

Solution:

x^{2} – 2x + 4 = 0

**2y**^{2}+ 6y = 3

Solution:

2y^{2} + 6y – 3 = 0

**15m**^{2}– 11m + 2 = 0

Solution:

15m^{2} – 11m + 2 = 0

**8r**^{2}= r + 2

Solution:

8r^{2} – r – 2 = 0

**(2x + 3)(3x – 2) + 2 = 0**

Solution:

(2x + 3)(3x – 2) + 2 = 0

6x^{2} – 4x + 9x – 6 + 2 = 0

6x^{2} + 5x – 4 = 0

**a(x**^{2}+ 1) = x(a^{2}+ 1)

Solution:

a(x^{2} + 1) = x(a^{2 }+ 1)

ax^{2} + a = a^{2}x + x

ax^{2} – a^{2}x – x + a = 0

ax^{2} –x(a^{2} + 1) + a = 0

**x**^{2}+ 8x + 6 = 0

Solution:

x^{2} + 8x + 6 = 0

Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic equations – Exercise 9.3 – Class X

Quadratic Equations – Exercise 9.4 – Class X