Completing the Square

The constant that completes a square

You already know from the squares-of-binomials chapter that squaring a sum produces a three-term pattern:

Read this from right to left and it says something useful. Any expression shaped like $x^2 + (\text{something})x + (\text{something})$ might be a squared binomial. Such an expression is a squared binomial only when its three pieces line up with $x^2$, $2px$, and $p^2$ for a single value of $p$. That matching is the whole idea of completing the square, so look closely at what it forces.

Start with just the first two terms, $x^2 + bx$. Compare the middle term $bx$ with the pattern’s middle term $2px$. For them to be the same, the coefficients must match, so $2p = b$, which means $p = \tfrac{b}{2}$. Half of the linear coefficient is the only value of $p$ that works. And once $p$ is pinned down, the pattern also tells you the missing last term: it must be $p^2 = \left(\tfrac{b}{2}\right)^2$.

So the recipe for the missing constant is short: take the coefficient of $x$, halve it, and square the result. For $x^2 + 10x$ the half is $5$ and the constant is $25$, giving $x^2 + 10x + 25 = (x + 5)^2$. For $x^2 - 6x$ the half is $-3$ and the constant is $9$, giving $x^2 - 6x + 9 = (x - 3)^2$. The sign inside the binomial follows the sign of $b$, because $p = \tfrac{b}{2}$ carries that sign along.

Rewriting a quadratic in completed-square form

A full quadratic $x^2 + bx + c$ usually does not come with the constant $\left(\tfrac{b}{2}\right)^2$ already in place; it comes with some other constant $c$. The trick is to conjure the constant you want without changing the expression, by adding it and subtracting it in the same breath. Adding and subtracting the same number changes nothing, since the two moves cancel, but it lets you regroup:

The braced part collapses to $\left(x + \tfrac{b}{2}\right)^2$, and whatever is left over stays outside as a plain constant:

This is the completed-square form. It says the same thing as the original quadratic, but now $x$ appears only once, safely inside the square. Written more generally, every quadratic can be put in the shape $(x - h)^2 + k$ for some constants $h$ and $k$, a form we will lean on again later.

There is a picture behind the name. The two terms $x^2 + bx$ tile an L-shape: a square of area $x^2$ and two strips whose areas together make $bx$. That L is missing one corner, and the corner you need to fill it is exactly a $\tfrac{b}{2}$ by $\tfrac{b}{2}$ square, of area $\left(\tfrac{b}{2}\right)^2$. Adding it literally completes the figure into a full square of side $x + \tfrac{b}{2}$.

Solving a quadratic by completing the square

Rewriting is the hard part; solving is what it unlocks. To solve $x^2 + bx + c = 0$, isolate the square and then take a square root. The steps run in a fixed order.

First move the constant to the right side, so the left holds only the $x$ terms:

Next add $\left(\tfrac{b}{2}\right)^2$ to both sides. Adding it to just the left would change the equation; adding it to both sides keeps the balance while turning the left into a perfect square:

The left side is now $\left(x + \tfrac{b}{2}\right)^2$. Take the square root of both sides, and here is the step people forget. Since the square root of a positive number splits into two cases, positive and negative, a $\pm$ sign appears:

Finally subtract $\tfrac{b}{2}$ to isolate $x$, which gives the general result

You do not need to memorize that line; it is faster and safer to run the four steps each time. What it shows is that completing the square always finishes a quadratic, whatever $b$ and $c$ are.

When $b$ is odd, half of it is a fraction, and $\left(\tfrac{b}{2}\right)^2$ becomes a fraction with $4$ in the denominator. Nothing about the method changes; you just carry the fractions carefully.

Real, repeated, and complex roots

Every solve reaches the same crossroads: after isolating the square you have

The number $D$ on the right decides everything, because you are about to take its square root, and there are exactly three cases.

If $D > 0$, its square root is a nonzero real number, so the $\pm$ produces two different real solutions, as in $x = 3 \pm \sqrt{5}$ above.

If $D = 0$, its square root is $0$, and $\pm 0$ is just $0$, so the two solutions collapse into one repeated solution $x = -\tfrac{b}{2}$. For instance $x^2 - 8x + 16 = 0$ becomes $(x - 4)^2 = 0$, whose only root is $x = 4$, met twice.

If $D < 0$, you are taking the square root of a negative number. Before this chapter that meant no solution, but you now have the imaginary unit $i$ with $i^2 = -1$, and $\sqrt{-1} = i$. So the root exists after all, and the $\pm$ produces two complex solutions, a conjugate pair $x = -\tfrac{b}{2} \pm (\text{something})\,i$.

This is the payoff the chapter was built for. Back when you only had real numbers, a quadratic with $D < 0$ simply had no answer, and you had to stop. Now that $i$ is available, the square root of a negative number is an ordinary complex number, so every quadratic has solutions, whether real or complex. That is exactly why this chapter introduced the imaginary unit before turning to quadratics: completing the square would otherwise dead-end on any equation with $D < 0$.

When the leading coefficient is not 1

Everything so far assumed the $x^2$ term stood alone, with coefficient $1$. When a quadratic starts with $ax^2$ for some $a \ne 1$, the pattern $x^2 + bx + \left(\tfrac{b}{2}\right)^2$ no longer matches directly. The fix is to make the leading coefficient $1$ first. For an equation, the cleanest way is to divide every term by $a$; because the right side is $0$, dividing does not disturb it. Then complete the square on the result as usual.

Running these same steps on the fully general quadratic $ax^2 + bx + c = 0$, with letters in place of numbers, produces a single formula for the roots. That is the subject of the next lesson; completing the square is the derivation behind it.

Common mistakes