Completing the Square

The identity that builds a perfect square

Everything here runs on one line of algebra you already know, read in an unfamiliar direction. Squaring a binomial gives

Read left to right, this expands a square. Read right to left, it is a recipe for recognizing one. A trinomial $x^2 + 2px + p^2$ is a perfect square, and the number multiplying $x$ in the middle is exactly twice the number $p$ that appears in the answer. That single observation tells you what constant a bare $x^2 + bx$ is missing.

Try it on a number before doing it in general. Take $x^2 + 6x$. If this is missing one constant $d$ to become a perfect square $(x + p)^2$, matching the middle terms means $2p = 6$, so $p = 3$, and matching the constants means $d = p^2 = 9$. Check it by expanding: $(x + 3)^2 = x^2 + 6x + 9$, which is exactly $x^2 + 6x$ with $d = 9$ added. Nothing about that reasoning depended on $6$ being the coefficient; the same matching works for any coefficient in place of $6$, which is what the argument below shows.

This is why the familiar instruction “take half the coefficient of $x$, then square it” is true. It is not a rule to memorize; it is a consequence of matching the middle term $2px = bx$, which forces $p = \tfrac{b}{2}$, and then reading off the constant $p^2$. Notice the added constant $\left(\tfrac{b}{2}\right)^2$ is never negative, whatever the sign of $b$, because squaring kills the sign.

The added constant almost never happens to be sitting there already. So we add it and immediately subtract it, changing the expression’s appearance without changing its value:

That equation is the engine of the lesson. Every worked example below is this one move, applied and then cleaned up.

Completing the square, literally

The name is not a metaphor. Picture $x^2 + bx$ as area. The $x^2$ is a square with side $x$. The $bx$ is a rectangle with one side $x$ and the other side $b$, so it is $b$ tall and $x$ wide, sitting next to the square. Now cut that rectangle down its length into two equal strips, each of width $\tfrac{b}{2}$ and length $x$. Glue one strip to the right edge of the square and the other to the bottom edge. The two strips fit flush, and what you have built is almost a bigger square of side $x + \tfrac{b}{2}$, missing only the little corner where the two strips would meet. That corner is a square of side $\tfrac{b}{2}$, so its area is $\left(\tfrac{b}{2}\right)^2$. Fill it in and the figure is a complete square of side $x + \tfrac{b}{2}$.

The picture draws both $x$ and $b$ as positive lengths, since an actual rectangle cannot have a negative side. But the identity it illustrates is pure algebra, and algebra does not care about the sign of $x$ or $b$. The very next worked example runs this same move on $x^2 - 6x$, where $b = -6$: the picture stops making sense as a picture, but the equation $x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$ still holds, because it was already proved above for every real number $b$, whatever real number $x$ eventually turns out to be too.

The area bookkeeping is the algebra. The original pieces have total area $x^2$ (the big square) plus $2 \cdot \tfrac{b}{2} \cdot x = bx$ (the two strips), which is $x^2 + bx$, our starting expression. Fill in the corner of area $\left(\tfrac{b}{2}\right)^2$ and the whole figure is a square of side $x + \tfrac{b}{2}$:

That is the identity from the last section, now seen rather than derived. The constant you add is the geometric corner that was missing, and “completing the square” describes precisely what your hand does to the picture.

The algorithm: standard form to a single square

To rewrite a whole quadratic, complete the square on its $x^2 + bx$ part and carry the leftover constant along. Start with a monic quadratic, one whose leading coefficient is $1$.

When the leading coefficient is not $1$, the square you are completing lives inside a factor of $a$, so pull $a$ out of the $x^2$ and $x$ terms first. This is the step most often botched, because the $a$ has to come back out at the end.

Two payoffs from one computation

Completing the square is the bridge from standard form to vertex form. The very same finished expression then answers two different questions depending on what you do with it next. Convert and read, and you get the vertex. Convert and solve, and you get the roots. One computation, two payoffs.

Payoff one: the vertex, the extremum, and the range

Run the algorithm on a completely general quadratic and the vertex appears in closed form.

Look at the $x$-coordinate: $-\dfrac{b}{2a}$ is exactly the axis of symmetry the chapter derived earlier from the parabola’s mirror symmetry. That lesson found the vertex’s height too, but only by a separate evaluation, plugging the axis back into $f$. Completing the square reaches the same height, $c - \dfrac{b^2}{4a}$, which is exactly $f\!\left(-\tfrac{b}{2a}\right)$, without a separate evaluation step: both coordinates fall out of the one rewritten expression, $x = h$ and $y = k$ together. From the vertex form the extremum and range follow with no extra work: when $a > 0$ the vertex is the minimum and the range is $y \ge k$. When $a < 0$ the vertex is instead the maximum and the range is $y \le k$. That is the whole reason vertex form is worth reaching, and completing the square always reaches it.

Payoff two: solving by isolating the square

To solve instead of read, do not distribute anything: get the square by itself on one side, then undo it. Undoing a square is where a genuinely important subtlety lives, so look at a number before stating it in general.

Take $x^2 = 9$. Two numbers square to $9$: $3^2 = 9$ and $(-3)^2 = 9$, so both $x = 3$ and $x = -3$ solve the equation. Yet the symbol $\sqrt{9}$ names only one number, $3$, the nonnegative one. If you solved $x^2 = 9$ by writing $x = \sqrt{9} = 3$, you would report a true solution but miss the other one, $x = -3$, which is exactly as valid. The general case, $x^2 = k$ for any $k \ge 0$, works the same way, and now you can see why.

So the $\pm$ is not decoration, it is the second root. Writing $x = \sqrt{k}$ by itself silently throws away $x = -\sqrt{k}$, and that lost solution is one of the most common errors in all of algebra. One boundary case is worth seeing directly: if the isolated square already equals $0$, such as $(x - 4)^2 = 0$, then $4 + \sqrt{0}$ and $4 - \sqrt{0}$ are both just $4$, so the two solutions collapse into one, $x = 4$. With that pinned down, solving is mechanical: complete the square, isolate it, and extract the root with a $\pm$.

Do this same isolate-and-extract computation on the fully general $ax^2 + bx + c = 0$, carrying the letters instead of numbers, and the answer you get has a name: the quadratic formula. That is the next lesson, and it is why the formula is a theorem you can prove rather than a spell to memorize. You already own its one idea.

Common mistakes