Completing the Square

Intuition

The last lesson solved quadratics by factoring, and factoring is a search: you hunt for two integers with the right sum and product, and sometimes there are none. Try $x^2 - 6x + 7 = 0$ and the hunt comes up empty, because no pair of rational numbers works, yet the parabola plainly crosses the axis twice. So factoring cannot be the whole story. This lesson builds the method that never fails. It takes any quadratic in standard form and rewrites it, with no guessing, as one perfect square plus a constant. Written that way, the quadratic is in exactly the vertex form from the start of the chapter. That single rewrite is the workhorse of the whole chapter: it hands you the vertex without a graph, and it solves the equation without a factorization. The name comes from a picture, an actual square with a corner missing that you fill in, and that picture is where we begin.

Learning goals

Add $\left(\tfrac{b}{2}\right)^2$ to complete the square
Factor $a$ out first, then distribute it back
Derive general vertex form and read the vertex
Solve from $(x + p)^2 = k$ with a $\pm$
Prefer it to factoring, since it never fails

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

(x + p)^2 = x^2 + 2px + p^2.

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.

The one constant that completes a perfect square#

Fix a number $b$ . We ask which constant $d$ makes $x^2 + bx + d$ a perfect square, meaning equal to $(x + p)^2$ for some $p$ .

Suppose $x^2 + bx + d = (x + p)^2 = x^2 + 2px + p^2$ . Two polynomials are equal only when their matching coefficients agree, so the $x$ terms give $b = 2p$ and the constants give $d = p^2$ . The first equation forces $p = \dfrac{b}{2}$ , and then the second forces

d = p^2 = \left(\frac{b}{2}\right)^2.

Conversely, if $d = \left(\dfrac{b}{2}\right)^2$ , the identity runs forward: $x^2 + bx + \left(\dfrac{b}{2}\right)^2 = \left(x + \dfrac{b}{2}\right)^2$ . So the constant that completes the square is not just a choice, it is the only choice, and it equals $\left(\dfrac{b}{2}\right)^2$ .

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:

x^2 + bx = \underbrace{x^2 + bx + \left(\tfrac{b}{2}\right)^2}_{\text{a perfect square}} - \left(\tfrac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2.

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 $x + \tfrac{b}{2}$ .

Completing the square for x squared plus b x. The large square is x by x. The b by x rectangle has been cut into two strips of width b over 2 and glued to the right and bottom edges. The only piece missing from the full square of side x plus b over 2 is the small accented corner of area (b over 2) squared. Adding that corner turns the shape into a perfect square.

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 $x + \tfrac{b}{2}$ :

x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^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 $1$ .

Worked example 1 A monic quadratic to vertex form

Rewrite $x^2 - 6x + 7$ as a perfect square plus a constant.

The variable part is $x^2 - 6x$ , so here $b = -6$ and half of it is $-3$ . The completing constant is $(-3)^2 = 9$ . Add it and subtract it so the value is untouched:

x^2 - 6x + 7 = \big(x^2 - 6x + 9\big) - 9 + 7.

The bracketed trinomial is the perfect square $(x - 3)^2$ , and the loose constants combine to $-9 + 7 = -2$ :

x^2 - 6x + 7 = (x - 3)^2 - 2.

This is vertex form $a(x - h)^2 + k$ with $a = 1$ , $h = 3$ , $k = -2$ . No factoring, no guessing, and it worked on the very quadratic that had no rational factorization.

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.

Worked example 2 A leading coefficient, and reading the vertex

Write $f(x) = 2x^2 + 12x + 7$ in vertex form, then read off its vertex, its minimum value, and its range.

Factor $2$ out of the first two terms only. The constant $7$ stays outside, because it has no $x$ in it:

2x^2 + 12x + 7 = 2\big(x^2 + 6x\big) + 7.

Complete the square inside the parentheses. There $b = 6$ , half is $3$ , and $3^2 = 9$ , so add and subtract $9$ inside:

2\big(x^2 + 6x + 9 - 9\big) + 7 = 2\big((x + 3)^2 - 9\big) + 7.

Now distribute the $2$ back across both pieces, the trap being to distribute it only to the square. The $-9$ is inside the parentheses, so it is multiplied by $2$ as well:

2(x + 3)^2 - 18 + 7 = 2(x + 3)^2 - 11.

In vertex form $f(x) = 2(x + 3)^2 - 11$ , so the vertex is $(-3, -11)$ . Because $a = 2 > 0$ the parabola opens upward, the vertex is the lowest point, the minimum value is $-11$ , and the range is $y \ge -11$ .

Check your understanding

Written in vertex form, $x^2 + 8x + 3$ equals which of these?

(x + 4)^2 - 13

(x + 4)^2 + 3

(x - 4)^2 - 13

(x + 4)^2 - 19

Two payoffs from one computation

Here is the point of the chapter. The three forms of a quadratic each answer a different question for free, and completing the square is the bridge from standard form to vertex form. But the very same finished expression 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.

Completing the square gives the vertex of every parabola#

Take $f(x) = ax^2 + bx + c$ with $a \ne 0$ . Factor $a$ from the first two terms:

f(x) = a\left(x^2 + \frac{b}{a}x\right) + c.

Inside, the coefficient of $x$ is $\tfrac{b}{a}$ , so half of it is $\tfrac{b}{2a}$ and the completing constant is $\left(\tfrac{b}{2a}\right)^2 = \tfrac{b^2}{4a^2}$ . Add and subtract it inside:

f(x) = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c = a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right) + c.

Distribute $a$ across both terms in the outer parentheses. The second term becomes $a \cdot \tfrac{b^2}{4a^2} = \tfrac{b^2}{4a}$ :

f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c.

This is vertex form $a(x - h)^2 + k$ with $h = -\dfrac{b}{2a}$ and $k = c - \dfrac{b^2}{4a}$ , so the vertex is

\left(-\frac{b}{2a},\ \, c - \frac{b^2}{4a}\right).

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 could locate the axis but not the height of the vertex. Completing the square supplies the missing $y$ -coordinate, $c - \dfrac{b^2}{4a}$ , which is just $f\!\left(-\tfrac{b}{2a}\right)$ evaluated once and for all. 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.

Check your understanding

What is the minimum value of $f(x) = x^2 - 10x + 30$ ?

30

5

-5

25

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 state it exactly.

Why the square root carries a plus or minus#

The symbol $\sqrt{k}$ denotes, by definition, the nonnegative number whose square is $k$ (this needs $k \ge 0$ ). So $\sqrt{k}$ is one specific number, never two. The equation $x^2 = k$ is a different object: it asks for every $x$ whose square is $k$ .

For $k > 0$ , move everything to one side and factor as a difference of squares:

x^2 = k \iff x^2 - k = 0 \iff \big(x - \sqrt{k}\big)\big(x + \sqrt{k}\big) = 0.

By the Zero Product Property a product is zero exactly when a factor is zero, so $x = \sqrt{k}$ or $x = -\sqrt{k}$ , which we write together as $x = \pm\sqrt{k}$ . Both are genuine solutions, and there are no others. For $k = 0$ the two collapse into the single solution $x = 0$ . For $k < 0$ there is no real solution at all, because a real square is never negative.

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. With that pinned down, solving is mechanical: complete the square, isolate it, and extract the root with a $\pm$ .

Worked example 3 Solving the quadratic that would not factor

Solve $x^2 - 6x + 7 = 0$ , the equation from the opening that had no rational factorization.

From Worked Example 1 the left side is $(x - 3)^2 - 2$ , so the equation is

(x - 3)^2 - 2 = 0.

Isolate the square by adding $2$ to both sides:

(x - 3)^2 = 2.

The right side is positive, so take the square root of both sides and attach the $\pm$ :

x - 3 = \pm\sqrt{2}.

Finally add $3$ to both sides:

x = 3 \pm \sqrt{2}.

There are the two roots, $x = 3 + \sqrt{2}$ and $x = 3 - \sqrt{2}$ . They are irrational, which is exactly why the integer search of factoring could never have found them, and why a method that does not rely on guessing was needed.

Worked example 4 When the isolated square is negative

Solve $x^2 - 2x + 5 = 0$ .

Complete the square on $x^2 - 2x$ . Half of $-2$ is $-1$ , and $(-1)^2 = 1$ :

x^2 - 2x + 5 = (x^2 - 2x + 1) - 1 + 5 = (x - 1)^2 + 4.

So the equation is $(x - 1)^2 + 4 = 0$ . Isolate the square:

(x - 1)^2 = -4.

The left side is a square, so it is never negative for any real $x$ , while the right side is $-4$ . No real number satisfies this, so the equation has no real solution. Completing the square does not fail here; it succeeds, and what it reveals is that this parabola never touches the $x$ -axis. (Numbers whose square is negative are the subject of a later chapter, and they will give this equation two solutions of a new kind.)

Check your understanding

How many real solutions does $x^2 + 2x + 6 = 0$ have?

A two B one C none D infinitely many

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

Dropping the leading coefficient. For $2x^2 + 12x + 7$ you must factor the $2$ out first, $2(x^2 + 6x) + 7$ , and then multiply it back across both pieces when you finish. Completing the square on $2x^2 + 12x$ as if it read $x^2 + 12x$ adds the wrong constant and gives the wrong vertex.
Distributing $a$ to only the square. After $2\big((x + 3)^2 - 9\big) + 7$ , the $-9$ is inside the parentheses too, so it becomes $-18$ , giving $2(x + 3)^2 - 11$ , not $2(x + 3)^2 - 2$ .
Dropping the $\pm$ . From $(x - 3)^2 = 2$ the solution is $x = 3 \pm \sqrt{2}$ , two values. Writing only $x = 3 + \sqrt{2}$ throws away the root $3 - \sqrt{2}$ , because $\sqrt{2}$ by definition names only the nonnegative root.
Adding without subtracting. Completing the square must not change the value. Add $\left(\tfrac{b}{2}\right)^2$ and subtract it in the same line. When you are solving an equation you may instead add it to both sides, but you cannot add it to one side and forget the other.
Calling a negative right side an error. If the isolated square equals a negative number, the work is correct and the conclusion is “no real solution,” not a mistake to hunt down. A real square is never negative.
Using half of $b$ instead of half, squared. The completing constant is $\left(\tfrac{b}{2}\right)^2$ , not $\tfrac{b}{2}$ . For $x^2 + 6x$ you add $9$ , not $3$ .

Key takeaways

One identity drives everything: $x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$ . The constant that completes the square is forced to be $\left(\tfrac{b}{2}\right)^2$ , the “half, then square” rule, and it is the missing corner of the area picture.
The algorithm: to rewrite $ax^2 + bx + c$ , factor $a$ from the $x^2$ and $x$ terms, add and subtract $\left(\tfrac{b}{2a}\right)^2$ inside, then distribute $a$ back. Unlike factoring, it never fails.
General vertex form: $ax^2 + bx + c = a\left(x + \tfrac{b}{2a}\right)^2 + \left(c - \tfrac{b^2}{4a}\right)$ , so the vertex is $\left(-\tfrac{b}{2a},\ c - \tfrac{b^2}{4a}\right)$ . This confirms the axis of symmetry $x = -\tfrac{b}{2a}$ and supplies the vertex’s height.
Two payoffs, one computation. Read the vertex form for the vertex, extremum, and range; or isolate the square to solve.
Solving: from $(x + p)^2 = k$ , there are two real solutions $x = -p \pm \sqrt{k}$ when $k > 0$ , exactly one when $k = 0$ , and none when $k < 0$ .
The $\pm$ is a root, not a flourish. $\sqrt{k}$ is the nonnegative root only, so $x^2 = k$ has solutions $\pm\sqrt{k}$ ; leaving off the minus loses a solution.

Completing the Square

Learning goals

The identity that builds a perfect square

Completing the square, literally

The algorithm: standard form to a single square

Two payoffs from one computation

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

Payoff two: solving by isolating the square

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)