Completing the Square — Visual Intuition
By: Adam Hrankowski
Tags: Math, Mathematics Education, High School

PRECALCULUS

Completing the Square — a Visual Intuition

Wherein we discover the roots of the quadratic formula

You want to find the roots of a quadratic polynomial. Good news! The method of completing the square always works. What makes that so, and what other information can we glean from this method?

Let’s find out with an example. We’ll find the roots of:

Before we can even start, we need to do some pre-processing. First, that 5 at the beginning has got to go. We can vanquish it by dividing everything by 5.

This new easier-to-solve equation codes for a different parabola, but with the same roots.

The second thing we need to do is find a new home for the third term. We’ll move it over to the right-hand side of the equation. Remember, that when it crosses the floor (the = sign), it changes its allegiance (sign). Thus, it becomes +⅗.

This is what we want. The part on the left is the square that needs completing. It will always have the form, x² + bx, where b can be any number. The number on the right can be anything (even zero).

The above diagram represents x²+2x as the area of a rectangle. One side has length x; the other has length x + 2. We’ve divided the rectangle so it consists of a square and a smaller rectangle.

We next slice the smaller rectangle (area 2x) down the centre. We move half of the rectangle to the top of the diagram.

The clear portion of the diagram has the same area as the previous diagram. It is now a square with a bite missing in the top right corner. We’ve shaded in the bite to complete that square.

The square has side of width x + 1. Thus, its area is (x+1)². This area — less the shaded portion of area 1² — is the area of the original rectangle.

We want the (x+1)² all on its own. We’ll move the -1² to the right. It becomes +1²=+1.

To remove the parentheses, we take the square root of both sides. Both positive and negative solutions will work.

Finally, slide the +1 over to the right-hand side of the equation. This leaves us with the two roots.

This format shows us that vertical line down the centre of the curve is at x = -1. The roots are 2√2/5 to the left and right of this line.

We can generalize this process for any quadratic. Below, A, B and C can be any numbers:

As before, we need to remove the first coefficient. We divide everything by A.

Next, move the third term to the right. Change its sign.

The general diagram now looks like this. The only thing different is the coefficient before the x.

When we split the smaller rectangle as before, we cut that coefficient in half.

Our equation now takes the form: area of the square, less the shaded bit, equals the number on the right.

As before, move the second term (without the x) to the right (changing its sign). Take the square root of both sides.

Once we move the second term to the right, isolating the x, we have our roots.

We need only to clean up the bit inside the square root.

The denominator inside the square root is a perfect square. It will become 2A. Putting everything together, we arrive at the quadratic formula we all know and love.

Where in the first example, -1 revealed the vertical axis of symmetry, in the general equation, -B/2A plays that role.

Using the quadratic formula is the same as completing the square — on autopilot.