Equations
Completing the Square
Turning any quadratic into a perfect square plus a constant, the one rule that makes it work, and why the vertex falls out of the result for free.
What this page covers
- Complete the square on a monic quadratic
- Handle a leading coefficient other than 1
- Solve a quadratic by completing the square
- Read the vertex of a parabola directly from the completed form
The idea
A quadratic that happens to be a perfect square is trivial to solve: take the square root of both sides. Completing the square manufactures that situation for any quadratic, by adding whatever constant is needed and compensating.
Worked example — the source's first case
Solve x2 - 4x + 3 = 0.
Adding 1 to both sides turned x2 - 4x + 3 into x2 - 4x + 4, which is (x - 2)2.
(x - 3)(x - 1) = x2 - 4x + 3. The source verifies this way too.
The rule
To complete the square on x2 + bx, add (b2)2.
The source phrases it as a2 is half of (2a), squared — take half the coefficient of x, then square it.
The typed supplementary sheet instructs 'add half of b2 to both sides'. That would add 12b2; the correct quantity is (b2)2 = b24. For b = -8 the correct addition is 16, whereas half of b2 is 32. The worked examples on the same sheet use the correct quantity, so the wording is the error, not the arithmetic.
| Expression | Coefficient of x | Half it | Square it | Completed |
|---|---|---|---|---|
| x2 + 6x | 6 | 3 | 9 | (x + 3)2 - 9 |
| x2 - 8x | -8 | -4 | 16 | (x - 4)2 - 16 |
| x2 + 5x | 5 | 52 | 254 | (x + 52)2 - 254 |
| x2 + 2x | 2 | 1 | 1 | (x + 1)2 - 1 |
When the leading coefficient is not 1
Factor the leading coefficient out of the x2 and x terms only, complete the square inside the bracket, then expand back. The constant term stays outside throughout.
Worked example — 2x2 + 3x - 2 = 0
Adding (34)2 inside a bracket multiplied by 2 adds 2(34)2 to the left side, so that is what must be added on the right. Adding (34)2 to the right instead is the most common error in the non-monic case.
2(14) + 32 - 2 = 12 + 32 - 2 = 0, and 2(4) - 6 - 2 = 0. Both roots verify.
Worked example — 2x2 - 5x + 1 = 0
What the completed form tells you
Completing the square does more than solve. It rewrites the quadratic in vertex form, from which the turning point can be read directly.
The reasoning is simple: (x - h)2 is never negative and is zero exactly at x = h. So when a > 0 the smallest value of y is k, reached at x = h.
Worked example — the source's vertex case
Minimum y = -14 at x = 3. No calculus and no formula were needed; the answer is visible in the form.
The vertex of y = ax2 + bx + c sits at x = -b2a. Here --124 = 3, agreeing. That shortcut is itself derived by completing the square in general.
Deriving the quadratic formula
Completing the square on the general quadratic produces the formula, which is the deepest reason to learn the technique.
Everything in the formula now has a source. The -b2a is the vertex; the radical measures how far the roots sit either side of it; and b2 - 4ac decides whether they are real at all.
Common mistakes
| Mistake | Correct | Counterexample |
|---|---|---|
| Adding 12b2 | Add (b2)2 | b = -8: add 16, not 32 |
| Not compensating on the other side | Balance the equation | Adding to one side only changes the equation |
| Forgetting to scale the compensation when a ≠ 1 | Add a(b2a)2 outside | The bracket multiplies it |
| Dropping the ± at the square-root step | Both signs | (x - 2)2 = 1 gives x = 1 and x = 3 |
| Factoring a out of the constant term too | Only out of the x2 and x terms | Otherwise the compensation is wrong |
Frequently asked questions
What exactly do I add?
The square of half the coefficient of x. For x2 - 8x the coefficient is -8, half is -4, and its square is 16. Adding 12b2 instead of (b2)2 is a real and common error.
Why not just use the quadratic formula?
For finding roots, usually you would. Completing the square is what derives the formula, and it does something the formula does not: it produces the vertex form, which gives the maximum or minimum of the parabola directly.
What if the leading coefficient is not 1?
Factor it out of the x2 and x terms first, complete the square inside the bracket, then multiply back out. The source works 2x2 - 5x + 1 = 0 this way.
Does it work when the roots are complex?
Yes, and it shows why cleanly. (x + 2)2 = -1 has no real solution because a real square is never negative; over the complex numbers it gives x = -2 ± i.
Source. Handwritten teaching notes, Week 3, pages 4-5, and Week 4, page 1, with the typed supplementary notes on quadratics.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.