 #### How do you use geometry to solve quadratics?

To solve an equation of the form x² – axb, construct two lengths at right angles, ½a and √b. Form a right triangle and extend the hypotenuse by ½a. The new length, x, is the solution of the equation, which can be proved as follows.

Using the Pythagorean Theorem,
c² = (½a)² + (√b)² = ¼a² + b.

But c = x – ½a, so c² = (x – ½a)² = x² – a + ¼a.

Equating both forms of c², we get x² – a + ¼a = ¼a² + b.

Hence x² – axb = 0.

For example, to solve the equation x² – 6x – 8 So x = 3 + √17, this can be verified using an algebraic method. But note that it only provides one solution; an algebraic method would provide two. This is because the second solution is negative (3 – √17) and in a geometrical context it would be meaningless.

To solve an equation of the form x² + axb, as before, construct two lengths at right angles, ½a and √b. Form a right triangle and reduce the hypotenuse by ½a. The remaining length, x, will yield the solution to the equation. The reader may like to work through this proof for themselves, by reworking the previous result.

It will be noted that so far we have solved quadratics of the form x² – axb and x² + axb. So what of the other two forms, x² – ax + b and x² + ax + b?

It should become quite evident that x² + ax + b will never produce positive solutions, hence it will never have a geometrical equivalent. The other equation, however, lends itself to another construction method.

To solve an equation of the form x² – ax + b, construct the lengths ½a and √b at right angles. Produce a line parallel to the side ½a and perpendicular to the base, √b. Construct a circle centred at the top of the side ½a and radius ½a. The intersection of the circle with the perpendicular produces a segment which will solve the equation.

Let us prove this result. Using the Pythagorean Theorem,
c² = (½a)² – (√b)² = ¼a² – b.

And as c = x – ½a it follows that c² = x² – ax + ¼a².

Hence x² – ax + ¼a² = ¼a² – b, giving the desired result, x² – ax + b = 0.

Two things will be noted this time. Firstly the circle intersects the perpendicular twice and it turns out that the vertical height of the lower intersection is the second solution of the equation.

The second point is quite significant. It can be seen that the circle will only intersect the perpendicular, and so provide a solution, if the length ½a is equal to or exceeds the length √b. That is, ½a ł Öb giving the requirement ¼a² ł b, hence a² ł 4b or a² – 4b ł 0. This is a rather attractive geometrical demonstration of the conditions for a real solution to exist. The reader may like to revisit the previous constructions to establish necessary conditions for solutions to exist in those cases.

In the work of Descartes it appears that algebraic problems have an equivalence in the geometry and so the opposite would be implied. That is, geometrical problems have an algebraic equivalent. It was this insight that allowed mathematicians to successfully re-address the four Greek geometrical problems of antiquity: trisecting an angle, constructing a heptagon, doubling a cube (constructing a cube with a volume exactly twice a given cube) and squaring a circle (constructing a square with the same area as a given circle).