# Where does the quadratic equation come from?

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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:

ax2+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:

x2+(b/a)x+(c/a) =0   (eq 3)

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

x2+(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:

x2+(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)= 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[b2/4a2 - c] = sqrt[(b2 - 4ac)/4a2]   (eq 8)

Where in the last step I took out a factor of 1/4a2. We can now use that sqrt[AB] = sqrt[A]sqrt[B] to write:

sqrt[(b2 - 4ac)/4a2] = sqrt[1/4a2]sqrt[(b2 - 4ac)]   (eq 9)

But 1/4a2 is just (1/2a)2, so sqrt[1/4a2] = 1/2a !

So we have:

sqrt[1/4a2]sqrt[(b2 - 4ac)] = (1/2a)sqrt[(b2 - 4ac)]   (eq 10)

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

(x+(b/2a)) = +/- (1/2a)sqrt[(b2 - 4ac)]   (eq 11)

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

x = (-b/2a) +/- (1/2a)sqrt[b2 - 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!

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