Euclid's geometric solution of a quadratic equation

We find xx so that (ax)x=b2(a - x)x = b^{2}.

Draw a line of length aa.

Draw a line of length bb perpendicular to the line aa from its midpoint.

Draw a circle centred at the end of the line with radius 12a\large\frac{1}{2}\normalsize a.

Now xx is a solution of (ax)x=b2(a - x)x = b^{2}.

To see this use the fact that (12a)2(12ax)2=b2(\large\frac{1}{2}\normalsize a)^{2} -(\large\frac{1}{2}\normalsize a - x)^{2} = b^{2}.

But (12a)2(12ax)2=(ax)x(\large\frac{1}{2}\normalsize a)^{2} - (\large\frac{1}{2}\normalsize a - x)^{2} = (a - x)x so xx is the required solution.