Completing the square without memorizing steps

TL;DR:

• Completing the square rewrites x² + bx as a perfect square by adding (b/2)².

• Whatever you add to one side, you must add to the other, or you must subtract it again.

• The same move gives vertex form, so you can read off a parabola's turning point.

Most people meet completing the square as a recipe: halve the middle number, square it, add it to both sides. The recipe works, but it is easy to forget which number gets halved and why. This article rebuilds the recipe from a picture, so the steps stop being something to memorize.

A square with a corner missing

Take x² + 8x. Picture a square with side x, so its area is x². Now attach two rectangles to it, each x long and 4 wide. Each has area 4x, and together they add the 8x. The shape you get is a big square with one corner missing.

That missing corner is a 4 by 4 square, with area 16. Add it and the whole figure is a square of side x + 4. In algebra: x² + 8x + 16 = (x + 4)². The 4 came from splitting the 8 into two halves, and the 16 is that half squared.

Why you add the same amount to both sides

An equation is a balance. If you add 16 to the left side to complete the square, the right side needs 16 too, or the equation is no longer the one you started with. A second option is to add 16 and subtract 16 on the same side, which changes nothing in total. Both are the same idea: you may only do things that keep the value the same.

Worked example: x² + 8x + 3 = 0

1. Move the constant to the other side: x² + 8x = −3.

2. Halve the coefficient of x: 8 ÷ 2 = 4. Square it: 4² = 16.

3. Add 16 to both sides: x² + 8x + 16 = −3 + 16, which is (x + 4)² = 13.

4. Take the square root of both sides, keeping both signs: x + 4 = ±√13.

5. Subtract 4: x = −4 + √13 or x = −4 − √13.

Since √13 ≈ 3.606, the two solutions are about x ≈ −0.394 and x ≈ −7.606.

Check by substituting back

Use the completed form to check quickly. For x = −4 + √13, we get x + 4 = √13, and (x + 4)² = 13. The original expression is (x + 4)² − 16 + 3 = 13 − 16 + 3 = 0. It works. The other root has x + 4 = −√13, and squaring a negative gives the same 13, so it works too.

A numeric check works as well. Put x ≈ −0.394 into x² + 8x + 3: 0.155 − 3.152 + 3 = 0.003. The small leftover comes from rounding.

The bonus: vertex form

The completed form also tells you where the parabola turns. Start from y = x² + 8x + 3 and write it as y = (x + 4)² − 16 + 3, so y = (x + 4)² − 13. A square is never negative, so the smallest y occurs when x + 4 = 0. The vertex is (−4, −13).

Notice the vertex sits halfway between the two roots: the average of −4 + √13 and −4 − √13 is −4. That symmetry is why the method works for any parabola.

When the coefficient of x² is not 1

Divide the whole equation by that coefficient first, so the x² term is alone. For 2x² + 12x + 5 = 0, divide by 2 to get x² + 6x + 2.5 = 0. Then complete the square as before: add (6/2)² = 9 to both sides of x² + 6x = −2.5, giving (x + 3)² = 6.5.

Try one yourself: x² + 2x − 5 = 0

Do it on paper first. Move the constant across: x² + 2x = 5. Half of 2 is 1, and 1² = 1. Add 1 to both sides: (x + 1)² = 6. Take roots: x + 1 = ±√6, so x = −1 + √6 or x = −1 − √6, about 1.449 and −3.449.

The check uses the completed form again: (x + 1)² − 1 − 5 = 6 − 6 = 0. A numeric test agrees: 1.449² + 2 × 1.449 − 5 = 2.100 + 2.898 − 5 = −0.002, which is zero up to rounding.

Here the middle coefficient is positive, so the bracket is (x + 1) with a plus. The number inside the bracket is always half the middle coefficient, with its sign. If the middle term were negative, you would write (x − 1).

Common mistakes

• Adding the square to one side only.

• Halving the wrong number, such as halving the constant instead of the coefficient of x.

• Forgetting the ± when taking the square root, and losing one solution.

• Squaring the coefficient before halving it.

• Skipping the division when the leading coefficient is not 1.

Where you will use it next

Every quadratic formula you have seen comes from running this method on the general equation, as shown in why the quadratic formula works. It also rewrites circles. Take x² + y² + 6x − 4y − 12 = 0 and complete the square in both letters: the result is (x + 3)² + (y − 2)² = 25, a circle centred at (−3, 2) with radius 5. Check: expanding the circle form gives x² + 6x + 9 + y² − 4y + 4 = 25, which simplifies back to the original.

Frequently asked questions

Is completing the square always better than the quadratic formula?

No. The formula is faster for a one-off answer. Completing the square is better when you also want the vertex, or when you are rewriting a circle or a function for calculus.

What if the number on the right side is negative after the step?

Then the square equals a negative number, so there is no real solution. The parabola never reaches zero.