This lesson will cover the definition and the solution of a quadratic equation.

## Definition of a Quadratic Equation

An equation of the form ax^{2} + bx + c = 0 is known as a quadratic equation. Here, x is the variable and a, b and c are constants where a ≠ 0.

Why a ≠ 0? Well because if a = 0, then the equation no longer remains a quadratic equation. The defining feature of the quadratic equations is the x^{2} term. Similarly, in a cubic equation (ax^{3} + bx^{2} + cx + d = 0), the coefficient of x^{3} must be non-zero.

Moving on.

The values of the variable x that satisfy the equation are known as the **roots** of the equation. Every quadratic equation has exactly two roots. This is a consequence of the fundamental theorem of algebra.

In simpler terms, a quadratic expression can always be expressed as a product of two linear expressions. Equating the quadratic expression to zero is equivalent to equating this product to zero. This will always give us two roots.

I won’t be going into proving this theorem. You can search for the proof elsewhere. Let’s move on to the meaty part – solution of the quadratic equation.

## Solution of a Quadratic Equation

The solution of a quadratic equation involves factorizing it into two linear expressions. This will also kind of demonstrate the fundamental theorem of algebra.

I’ll take some special cases to begin with, and then move on to the general case.

**Case 1**: c = 0

In this case the quadratic equation looks something like this: ax^{2} + bx = 0. This can easily be factorized as x(ax + b) = 0.

Now, x(ax + b) = 0 implies that either x = 0 or ax + b = 0. And since a ≠ 0, the second linear equation gives the root x = – b/a.

Hence, we get two roots: **x = 0** and **x = – b/a.**

Let’s explore another simple case when b = 0.

**Case 2**: b = 0

The quadratic equation now looks like ax^{2} + c = 0.

This can be rearranged as x^{2} + c/a = 0 or x^{2} – \( (\sqrt{-c/a})^2 \)= 0, and now can be factorized as (x + \( \sqrt{-c/a} \)) (x – \( \sqrt{-c/a} \)) = 0.

Once again, we get two roots x = \( \sqrt{-c/a} \) and x = – \( \sqrt{-c/a} \).

In case c is also 0, then both the roots will be 0. (Let’s call this case **Case 0**.)

Don’t be scared by the negative sign inside the root. Also, you don’t have to remember any results (yet).

Let’s move on to the general case, where both b and c are non-zero.

**Case 3**: b ≠ 0 and c ≠ 0

Our aim again is to factorize the quadratic expression into two linear ones. Let’s begin.

Here’s the trick: if we’re able to use the first two terms of the equation to form a perfect square, the equation would look similar to that in Case 2.

I’ll first divide the equation by ‘a’ to clean it up. This gives us x^{2} + \( \frac{b}{a} \)x + \( \frac{c}{a} \) = 0.

Now, adding and subtracting \((\frac{b}{2a})^2 \) does the trick.

The equation becomes x^{2} + \( \frac{b}{a} \)x + \((\frac{b}{2a})^2 \) – \( (\frac{b}{2a})^2 \) + \( \frac{c}{a} \) = 0, or (x +\( \frac{b}{2a} \))^{2} + \( \frac{4ac-b^2}{4a^2} \) = 0.

This looks much like the one in Case 2, i.e. X^{2} + k = 0, where X = x +\( \frac{b}{2a} \) and k = \( \frac{4ac-b^2}{4a^2} \).

Our work is done. The roots are X = \( \frac{\sqrt{b^2-4ac}}{2a} \) and X = – \( \frac{\sqrt{b^2-4ac}}{2a} \).

This gives x + \( \frac{b}{2a} \) = ± \( \frac{\sqrt{b^2-4ac}}{2a} \) , or x = \( \frac{-b\pm\sqrt{b^2-4ac}}{2a} \).

This result, known as the **quadratic formula**, is something that you should remember.

I’ll stop here and continue in the next lesson with some examples.

## Lesson Summary

- An equation of the form ax
^{2}+ bx + c = 0, where a ≠ 0, is known as a quadratic equation. - Every quadratic equation has two roots. This is a consequence of the fundamental theorem of algebra.
- The roots of the equation ax
^{2}+ bx + c = 0 are given by x = \( \frac{-b\pm\sqrt{b^2-4ac}}{2a} \). This is known as the quadratic formula.