Solving a Quadratic with a Straightedge and Compass

In most high school algebra classes, students learn to solve a quadratic ax² + bx + c = 0 using a variety of different methods.  The most common methods taught are factoring, graphing, completing the square, or using the quadratic equation.  However, a method for solving a quadratic using a straightedge and compass is described near the end of the book Number: The Language of Science by Tobias Dantzig (first published in 1930).

To solve the quadratic equation ax² + bx + c = 0 using a straightedge and compass, let p = -^b/_a and q = ^c/_a.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(p, 0), Q(0, q), R(p, q), and S(p, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C, construct circle C with radius CU.  Circle C will then intersect the x-axis at the two solutions X₁ and X₂.

For example, let’s say you wanted to solve the quadratic x² – 6x + 5 = 0.  Then a = 1, b = -6, and c = 5, and p = -^b/_a = 6 and q = ^c/_a = 5.  Use a compass and straightedge to plot out the coordinates U(0, 1), P(6, 0), Q(0, 5), R(6, 5), and S(6, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(3, 3), construct circle C with radius CU.  Circle C intersects the x-axis at x = 1 and x = 5, which are the two solutions to x² – 6x + 5 = 0. 

We can show that this method will always work by finding a Cartesian equation for circle C, setting y equal to zero (for the x-axis), and simplifying the answer to being the quadratic equation (x = ^{(–b ± √(b^2 – 4ac))}/_2a).  The center of circle C is the midpoint of U(0, 1) and R(p, q), which is (^p/₂, ^{(q + 1)}/₂), and the radius of circle C is half the distance between U(0, 1) and R(p, q), which is r = ½√(p² + (q – 1)²).  Therefore, the equation for circle C is (x – ^p/₂)² + (y – ^{(q + 1)}/₂)² = (½√(p² + (q – 1)²))² or when y = 0, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²).  Applying some algebra, (x – ^p/₂)² + (^{(q + 1)}/₂)² = ¼(p² + (q – 1)²) è (2x – p)² + (q + 1)² = p² + (q – 1)² è (2x – p)² = p² + (q – 1)² – (q + 1)² è (2x – p)² = p² + (q² – 2q + 1) – (q² + 2q + 1) è (2x – p)² = p² – 4q è 2x – p = ±√(p² – 4q) è 2x = p ± √(p² – 4q) è x = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

If the quadratic equation ax² + bx + c = 0 has two imaginary solutions, circle C will not intersect the x-axis.  However, the solution can still be found with a straightedge and compass by performing a few additional steps.  First, construct a perpendicular bisector of QR through C called CM.  Second, construct a tangent OT of circle C.  Third, construct circle O with radius OT.  Circle O will then intersect line CM at the two solutions Z₁ and Z₂, in which the x-coordinate is the real part and the y-coordinate is the imaginary part. 

For example, let’s say you wanted to solve the quadratic x² – 4x + 5 = 0.  Then a = 1, b = -4, and c = 5, and p = -^b/_a = 4 and q = ^c/_a = 5.  As before, use a compass and straightedge to plot out the coordinates U(0, 1), P(4, 0), Q(0, 5), R(4, 5), and S(4, 1) and then construct rectangle UQRS and its diagonals.  Then, using where the diagonals of the rectangle intersect at point C(2, 3), construct circle C with radius CU.  This time, however, circle C does not intersect the x-axis, so construct a perpendicular bisector of QR through C called CM, construct a tangent OT of circle C, and construct circle O with radius OT.  Circle O intersects the line CM at Z₁(2, 1) and Z₂(2, -1), and so the two solutions are x = 2 ± i.

We can also show that these additional steps will always work by combining the Cartesian equation for circle O and the equation of the vertical line CM and showing that its solution (x, ±y) can be entered into x’ = x ± yi to simplify into the quadratic equation (x’ = ^{(–b ± √(b^2 – 4ac))}/_2a).  As mentioned above, the center of circle C is (^p/₂, ^{(q + 1)}/₂), and so the vertical line CM can be represented by x = ^p/₂.  The length of segment CO is the distance between O(0, 0) and C(^p/₂, ^{(q + 1)}/₂), which is CO = ½√(p² + (q + 1)²).  Segment CT is a radius of circle C, which we also know from above is CT = ½√(p² + (q – 1)²).  Since ΔCTO is a right angle triangle, CT² + OT² = CT², or substituting, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))².  Applying some algebra, (½√(p² + (q – 1)²))² + OT² = (½√(p² + (q + 1)²))² è OT² = ¼(p² + (q + 1)²) – ¼(p² + (q – 1)²) è OT² = ¼(p² + q² + 2q + 1) – ¼(p² + q² – 2q + 1) è OT² = ¼(4q) è OT² = q è OT = √q, which means the radius of circle O is √q.  The Cartesian equation for circle O is then x² + y² = q.  Combining this with x = ^p/₂, (^p/₂)² + y² = q è y² = q – (^p/₂)² è y² = ¼(4q – p²) è y = ±½√(4q – p²).  Now if x = ^p/₂ and y = ±½√(4q – p²), x’ = x + yi = ^p/₂ ± ½√(4q – p²)i = ^p/₂ ± ½√(4q – p²)√(-1) = ^p/₂ ± ½√(p² – 4q) = ^{(p ± √(p^2 – 4q))}/₂.  Substituting p = -^b/_a and q = ^c/_a as defined above, x’ = ^{(–b ± √(b^2 – 4ac))}/_2a, which is the quadratic equation.

Therefore, there exists a method for solving a quadratic with a straightedge and compass.  The steps themselves are fairly simple, but the proof is not.  Even more difficult than the proof must have been the invention of the method itself, which must have been by some creative genius.