# Deriving the Quadratic Formula

In this article, we will be looking at how we derive the Quadratic Formula from the general quadratic:

ax^{2} + bx + c = 0

The technique we will be using is known as completing the square. Typically, when we have a quadratic such as

x^{2} + 6x +9 = 0

We note that 6/2 = 3 and that 3^{2} = 9. How does that help us get to:

(x+3)^{2} = 0

Because if we expand and foil the left side of the equation, we have

(x+3)(x+3) = x^{2} + 3x + 3x + 9 = x^{2} + 6x +9

This is what we had initially. By dividing the middle term by two and then squaring it, we can rewrite quadratics into a simplified form that makes it much easier to solve for x. It is called completing the square because once applied, the simplified quadratic is something that is being squared, or raised to the 2^{nd} power. The middle term is a combination of two of the same terms, hence why I divide it by two to find the initial term that will be squared.

## A Problem

However, what if someone gave you the following?

3x^{2} + 7x -3 = 0

If we tried to complete the square like we did before, we would have the result

(x + 49/4)^{2} = 0

This would not be correct if foiled out. Completing the square works when a, the first term, is 1. This is so that I do not have any coefficient errors like the one above. So the first step would have been to divide everything by 3, to get

x^{2} + (7/3)x -1 = 0

Then move the -1 to the other side by adding it to both sides

x^{2} + (7/3)x = 1

Now we can use the completing the square technique, dividing the middle term and then squaring it, to get

(7/6)^{2} = (49/36)

Which we add to both sides to get

x^{2} + (7/3)x + (49/36) = 1 + (49/36) = (36/36) + (49/36) = (85/36)

Everything on the left is a perfect square, so that

(x + 7/6)^{2} = (85/36)

And now this is much like the first problem, solving for x using a bit of algebra. But how do I solve quadratics with any a, b, or c? It sure would be nice to just have a formula to plug in my values and find my values for x. That is what the derivation below will give us the ability to do.

## The Solution

Let us start with any quadratic equation with a, b and c coefficients.

ax^{2} + bx + c = 0

Divide both sides by a to get

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

Now it looks like quadratics with the x^{2} being by itself. Now we use the completing the square step, taking the middle term , dividing it by 2 and then squaring it. If I add this to one side, I need to do it to the other, and so we have

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

Rewriting this gives us

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

Now everything on the left is just like a typical completing the square problem, so

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

To check this, expand and foil

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

This is what we had initially. So, simplifying the left gives us

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

If we expand the term on the right, we get

(b/2a)^{2} - (c/a) = (b^{2}/4a^{2}) - (c/a)

To combine the two terms on the right, we need to have a common denominator. In this case, that would be . To go from a to , we need to multiply by 4a, so

(b^{2}/4a^{2}) - (c/a) = (b^{2}/4a^{2}) - (4ac/4a^{2}) = (b^{2} - 4ac)/4a^{2}

Now, since we are aiming to solve for x, we need to take the square root of both sides so that we may further isolate the variable. Doing do yields:

{[x + (b/2a)]^{2}}^{1/2} = [x + (b/2a)]

and

{(b^{2} - 4ac)/4a^{2}}^{1/2} = ±(b^{2} - 4ac)^{1/2}/2a

Thus

[x + (b/2a)] = ±(b^{2} - 4ac)^{1/2}/2a

It is ± because no matter what the sign is of the number you square, it will always be positive. Finally we subtract from both sides, so

x = -(b/2a) ±(b^{2} - 4ac)^{1/2}/2a = [-b ± (b^{2} - 4ac)^{1/2}]/2a

The Quadratic Formula!

- What Are Irrational Numbers?

To better understand irrational numbers, we need to know what a rational number is. This is simply a number that can be defined as a fraction of two real numbers. 5 is rational because 5/1 = 5. 1.6 is also rational because 16/10 = 1.6. Irrational... - Early Proofs of the Pythagorean Theorem

While scholars will argue about whether or not Pythagoras actually discovered the theorem that bears his name, it is still one of the most important theorems in mathematics. Evidence that the ancient Indians and Babylonians knew of its principles...

**© 2013 Leonard Kelley**

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