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)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!