Completing the Square to Solve

In this lesson you will: Add the right constant to turn x^2+bx into a perfect square; Transform any quadratic equation into the form (x-p)^2=q with the same solutions; Finish by taking square roots, and leave irrational solutions in exact form.

A method with a target

Factoring is fast when it works. It fails the moment the solutions are irrational, and most quadratics have irrational solutions.

Completing the square never fails, because it does not depend on the numbers being friendly. Ohio states the goal explicitly: transform the equation into (x-p)^2=q, an equation with the same solutions that Lesson 16.1 already knows how to finish.

The engine is one identity: x^2+bx+((b)/(2))^2=(x+(b)/(2))^2. Half of b, then squared, is the constant that completes the square.

The five moves

Take x^2+6x-7=0. Move the constant out of the way: x^2+6x=7. Half of 6 is 3, and 3^2=9, so add 9 to both sides: x^2+6x+9=16.

The left side is now (x+3)^2, so (x+3)^2=16. Square-root both sides: x+3=+/-4, giving x=1 or x=-7.

Those five moves — isolate, halve, square, add to both sides, factor — work on any quadratic in x.

When the leading coefficient is not 1

The identity is written for x^2 with no coefficient, so make that true first: divide every term by a.

2x^2+8x-10=0 becomes x^2+4x-5=0 after dividing by 2. Then x^2+4x=5, add 4, and (x+2)^2=9, so x=1 or x=-5.

Ohio keeps these tame: when the leading coefficient is not 1 it is one of -1, +/-2 or +/-3, and b divides evenly by it, so the division never produces a fraction.

Common mistake — Adding the new constant to only one side

Solving x^2+10x=3, a student writes x^2+10x+25=3, then (x+5)^2=3.

The attention is all on the left, where the interesting work is happening, and the phrase \"complete the square\" points at one expression rather than at an equation. So the 25 goes in and the right side is copied across untouched.

But adding 25 to one side of an equation changes what the equation says. The correct line is x^2+10x+25=3+25, that is (x+5)^2=28, giving x=-5+/-2sqrt(7).

The wrong version gives x=-5(3), about -3.27 and -6.73. Substituting -3.27 into x^2+10x gives about -22.0, nowhere near 3.

How to catch it: Expand your finished form and compare it with the original. (x+5)^2=28 expands to x^2+10x+25=28, which is x^2+10x=3. If the expansion does not return the equation you started with, the constant went on one side only.