How to Solve Quadratic Equations? There are basically three methods to solve quadratic equations. They are:

- Using Quadratic formula
- Factoring the quadratic equation
- Completing the square

A quadratic equation is an equation that has the highest degree equal to two. The standard form of the quadratic equation is ax^{2} + bx + c = 0, where a, b, c are constants and a ≠ b ≠ 0. Here, x is an unknown variable for which we need to find the solution. Let us learn here how to solve quadratic equations.

## Solving Quadratic Equations – Using Quadratic Formula

The quadratic formula is used to find solutions of quadratic equations. If ax^{2} + bx + c = 0, then solution can be evaluated using the formula given below;

\(\begin{array}{l}x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)

Thus, the formula will result in two solutions here.

\(\begin{array}{l}x_1 = \frac{-b- \sqrt{b^{2}-4ac}}{2a}\end{array} \)

or

\(\begin{array}{l}x_2 = \frac{-b+ \sqrt{b^{2}-4ac}}{2a}\end{array} \)

Here, b^{2} – 4ac is the discriminant (D).

D = b^{2} – 4ac

- If D = 0, the two roots of quadratic equation are real and equal
- If D > 0, the roots are real and unequal
- If D < 0, the roots are not real, i.e. imaginary

**Example: Solve x ^{2} – 5x + 6 = 0.**

Solution: Given,

x^{2} – 5x + 6 = 0

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

By the quadratic formula, we know;

\(\begin{array}{l}x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)

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

Thus, the roots are real.

Hence,

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

Therefore, the solution of x^{2} – 5x + 6 = 0 is 3 or 2.

## Solving Quadratic Equations – By Factorisation

We can write the quadratic equation as a product of factors having degree less than or equal to two. This method of solving quadratic equations is called factoring the quadratic equation.

Let us learn by an example.

**Example: Solve 6m ^{2} – 7m + 2 = 0 by factoring method.**

Solution: 6m^{2} – 4m – 3m + 2 = 0

⟹ 2m(3m – 2) – 1(3m – 2) = 0

⟹ (3m – 2) (2m – 1) = 0

⟹ 3m – 2 = 0 or 2m – 1 = 0

⟹ 3m = 2 or 2m = 1

Therefore, the solutions of the given equation are:

m = ⅔ or m = ½

## Solving Quadratic Equation – Completing Square

To solve the quadratic equation using completing the square method, follow the below given steps.

- First make sure the equation is in the standard form: ax
^{2}+ bx + c = 0 - Now, divide the whole equation by a, such that the coefficient of x
^{2}is 1. - Write the equation with a constant term on the Right side of equation
- Add the square of half of coefficient of x on both sides and complete the square
- Write the left side equation as a square term and solve

Let us understand with the help of an example.

Example: Solve 4x^{2} + x = 3 by completing the square method.

Solution: Given,

4x^{2} + x = 3

Divide the whole equation by 4.

x^{2} + x/4 = ¾

Coefficient of x is ¼

Half of ¼ = ⅛

Square of ⅛ = (⅛)^{2}

Add (⅛)^{2} on both sides of the equation.

x^{2} + x/4 + (⅛)^{2} = ¾ – (⅛)^{2}

x^{2} + x/4 + 1/64 = ¾ + 1/64

(x + ⅛)^{2} = (48+1)/64 = 49/64

Taking square root on both sides, we get;

x+1/8 = √(49/64) = ±7/8

x = ⅞ – ⅛ = 6/8 = ⅜

x = – ⅞ – ⅛ = -8/8 = -1

Hence, the solution of quadratic equation is x = ⅜ or x = -1.

## Practice Questions

Solve the following quadratic equations.

- 2s
^{2}+ 5s = 3 - 9x
^{2}– 6x – 2 = 0 - 3x
^{2}– 11x – 4 = 0 - 2x
^{2}– 12x – 9 = 0

## Frequently Asked Questions on Solving Quadratic Equations

Q1

### What are the methods to solve quadratic equations?

There are majorly three methods of solving quadratic equations. They are:

- Using Quadratic formula
- Factoring the quadratic equation
- Completing the square

Q2

### What is the formula to solve quadratic equations?

The quadratic formula is given by:

x = [b±√(b²-4ac))]/(2a)

Q3

### What is the discriminant of the quadratic formula?

The discriminant of quadratic formula is b2 – 4ac. It helps to determine the nature of the roots.

Q4

### When is the quadratic equation has imaginary roots?

When discriminant of the quadratic equation is less than zero, then the roots are imaginary or non-real.