Friday, January 1, 2010

Solving quadratic equations via geometric construction

We can solve a quadratic equation of the form x2 - sx + p = 0, s, p \in \R, using the standard construction tools of compass and straightedge.  The method has been attributed to critic Thomas Carlyle.

Construct the circle in the Cartesian plane with center and passing through A(0,1).  By symmetry, the circle also passes through B(s, \, p) and C(0,p).

Circle in Cartesian plane

Because the circle has center   and passes through A(0,1), the equation of the circle is

\left( x-\frac{s}{2}\right)^2 +\left( y-\frac{p+1}{2}\right)^2 = \left( \frac{s}{2}\right)^2 +\left( \frac{p+1}{2}-1\right)^2

This reduces to

x2 - sxy2 - (p + 1)y + p = 0

So the x-intercepts of the circle are the solutions to  x2 - sx + p = 0.

Alternate justification:

The segment joining the x-intercepts has a length x_2 - x_1 = 2\left(\frac{s}{2} - x_1 \right), hence x1 + x2 = s.

Circle in Cartesian plane

\angle OCX_2 intercepts the arc AX1X2, and\angle AX_1X_2  intercepts the opposite arc, hence the two angles are supplementary.  But \angle AX_1 X_2 is also supplementary with \angle O X_1 A, so \angle OC X_2 is congruent to \angle OX_1 A, which in turn impies that  \triangle COX_2 \sim \triangle X_1 OA.  Hence


which implies that \frac{x_1}{1}=\frac{p}{x_2}, so x1 x2 = p.

Thus (xx1)(x - x2) = x2 - sx + p, and the solutions to x2 - sx + p = 0 are x1 and x2 .


j edward ladenburger said...

you might want to change the y coordinate of your circle center -- shows as \frac{p-1}{2} when you want \frac{p+1}{2}

j edward ladenburger said...

you may want to change the y-coordinate of your circle center -- it reads \frac{p-1}{2} when you want \frac{p+1}{2}

Bruce Yoshiwara said...