# Where does the quadratic equation come from?

This is a quick derivation of the quadratic formula for an GCSE Maths students looking to stretch themselves a little and go for that A*!

If we have a general quadratic equation:

ax^{2}+bx+c=0 (eq1)

You will have been told that the solution is:

x= (-b +/- sqrt[b^2-4ac])/2a (eq 2)

But *why* is this true? Let's go through this in steps. Firstly, we will divide our equation by a:

x^{2}+(b/a)x+(c/a) =0 (eq 3)

Now we can use a clever trick called *'completing the square'*, which states that:

x^{2}+(something)x = (x+(something)/2)^{2} - (something/2)^{2 }(eq 4)

You can check this by multiplying out the brackets on the right hand side!

Anyway, this clever trick tells us that:

x^{2}+(b/a)x+(c/a) = (x+(b/2a))^{2} - (b/2a)^{2} + (c/a) = 0 (eq 5)

Let's tidy up by moving the contants over to the right hand side:

(x+(b/2a))^{2} = (b/2a)^{2} - (c/a) (eq 6)

Now we take the sqaure root of both sides:

(x+(b/2a)) = +/- sqrt[(b/2a)^{2} - (c/a)] (eq 7)

The reason for the +/- is that there are always two answers to the sqrt (for example, (2)^{2 }= 4, but so does (-2)^{2}, so sqrt[4] = +/- 2 ). Let's look at the square root on the right hand side and tidy it up a bit:

sqrt[(b/2a)^{2} - c] = sqrt[b^{2}/4a^{2} - c] = sqrt[(b^{2} - 4ac)/4a^{2}] (eq 8)

Where in the last step I took out a factor of 1/4a^{2}. We can now use that sqrt[AB] = sqrt[A]sqrt[B] to write:

sqrt[(b^{2} - 4ac)/4a^{2}] = sqrt[1/4a^{2}]sqrt[(b^{2} - 4ac)] (eq 9)

But 1/4a^{2} is just (1/2a)^{2}, so sqrt[1/4a^{2}] = 1/2a !

So we have:

sqrt[1/4a^{2}]sqrt[(b^{2} - 4ac)] = (1/2a)sqrt[(b^{2} - 4ac)] (eq 10)

Phew, we're nearly there now! Using this into equation (7) we have:

(x+(b/2a)) = +/- (1/2a)sqrt[(b^{2} - 4ac)] (eq 11)

Let's subtract (b/2a) from both sides to make x the subject:

x = (-b/2a) +/- (1/2a)sqrt[b^{2} - 4ac] (eq 12)

And now (last step!) we take out a factor of (1/2a) and we have at last:

x = (-b +/- sqrt[b^2-4ac])/2a (eq 2)

And we're done!