In the article on Solving Quadratics you were introduced to the Quadratic Formula as a way to solve all quadratic equations. In this extra article we’ll take a look at how this formula was derived.

Note – This is an extra article for those who are interested in how we got to the quadratic formula, you do not need to know this for your exam!

We have our general quadratic in the form:

$ax^2 + bx + c = 0$

We are now going to Complete the Square to deduce the quadratic formula. You should remember from the Completing the Square section in the Solving Quadratics article, that we first need to get the $x^2$ on it’s own, so we divide by $a$

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

The next thing we need to do is to move the number in it’s own to the left hand side of our equation, giving us

$x^2 + \frac{b}{a}x = -\frac{c}{a}$

Now this is probably the trickiest bit of the whole operation. We take our $x$ plus half the coefficient of our original $x$, and square it. Remembering to subtract half the coefficient squared.

$(x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} = -\frac{c}{a}$

If you can’t see how I got to this point, I advise you to take a look at the article mention before, where I go through the steps in more detail.

Now adding $- \frac{b^2}{4a^2}$ to both sides gives us

$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} -\frac{c}{a}$

What we want now is to combine these two fractions on the left hand side. To do that we need to get them both over a common denominator, so multiplying the numerator and denominator of the last fraction by $4a$ gives us

$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} -\frac{4ac}{4a^2}$

The left hand side can now be simplified.

$(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$

Now taking the square root of both sides gives us

$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$

We need to remember here that taking the square root gives you both a positive and negative result. The left hand side can again be simplified to

$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

We now subtract the $\frac{b}{2a}$ from both sides to give a value for $x$

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

As both the fractions on the left hand side have common denominator, they can again be simplfied.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Congratulations, you have just proved the Quadratic Formula!

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

I must add again that this is not needed for your exam, nor do you need to remember the formula as it is given you. This is just a little extra article for those with an interest in maths.