Completing the Square

Intuition

Factoring solves a quadratic fast when it splits into tidy whole-number pieces, but most quadratics do not split that way, and then factoring gets stuck. Take $x^2 + 6x$ . Factoring pulls out $x$ , giving $x(x + 6)$ , but that is not a perfect square, and no factor pair turns it into one. One particular constant, added in, does. Completing the square is the method that finds that constant, and unlike factoring, it always works, on any quadratic. The trouble with a general equation $x^2 + bx + c = 0$ is that the unknown $x$ shows up in two places, the $x^2$ and the $bx$ , so you cannot isolate it directly. The fix is to gather those two terms into a single perfect square $(x + \text{something})^2$ plus a leftover constant. Once $x$ appears only once, inside that square, taking a square root finishes the job. The same move works on any quadratic, and in the next lesson it becomes the derivation behind a single formula.

Learning goals

Add the square of half the linear coefficient
Rewrite a quadratic in completed-square form
Keep both signs when taking a square root of both sides
Sort the roots by the sign of the number under the square root
Divide by $a$ first when the leading coefficient is not one

The constant that completes a square

Take $x^2 + 6x$ . What constant turns it into a perfect square? The picture below shows the answer. The two terms tile an L-shape, a square of area $x^2$ and two strips whose areas together make $6x$ . That L is missing one corner, and the corner that fills it is $3$ by $3$ , area $9$ , because each strip already has length $3$ . Adding that $9$ completes the figure into a full square of side $x + 3$ :

x^2 + 6x + 9 = (x + 3)^2.

Completing the square, seen as area. The piece of area x squared and the two strips of area 3x each tile an L-shape. Adding the 3-by-3 corner of area 9 fills the notch, making a full square of side x + 3, so x squared plus 6x plus 9 equals (x + 3) squared.

The picture works here because $6$ is positive, so each strip has a sensible length. The pattern behind it is algebraic, and algebra also covers a negative middle coefficient, where there is no picture to draw. You already know from the squares-of-binomials lesson that squaring a sum produces a three-term pattern:

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

Read this from right to left. Compare $x^2 + 6x$ with $x^2 + 2px + p^2$ : the middle terms agree when $2p = 6$ , so $p = 3$ , and the last term must be $p^2 = 9$ , exactly the corner you just added. The same matching works for any linear coefficient $b$ in place of $6$ . Compare $x^2 + bx$ with $x^2 + 2px + p^2$ : the middle terms agree when $2p = b$ , so $p = \tfrac{b}{2}$ , and the constant that completes the square is $p^2 = \left(\tfrac{b}{2}\right)^2$ , the square of half the linear coefficient.

A quick check confirms it. Squaring $x + \tfrac{b}{2}$ gives $x^2 + 2\cdot\tfrac{b}{2}\cdot x + \left(\tfrac{b}{2}\right)^2$ , and $2\cdot\tfrac{b}{2} = b$ , so that is exactly $x^2 + bx + \left(\tfrac{b}{2}\right)^2$ , the trinomial you want.

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.

Check your understanding

What constant must you add to $x^2 + 12x$ to complete the square?

36

6

24

144

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 fix is to add the constant you want and immediately subtract it again. Adding and subtracting the same number changes nothing, since the two moves cancel, but it lets you regroup:

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

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

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

This is the completed-square form. It says the same thing as the original quadratic, but now $x$ appears only once, inside the square. Written more generally, every quadratic with leading coefficient $1$ can be put in the shape $(x - h)^2 + k$ for some constants $h$ and $k$ . This is a form we will lean on again later.

Worked example 1 Rewrite $x^2 + 8x + 3$ in completed-square form

The linear coefficient is $b = 8$ , so half of it is $4$ and the constant that completes the square is $4^2 = 16$ . Add and subtract that $16$ :

x^2 + 8x + 3 = \left(x^2 + 8x + 16\right) - 16 + 3.

The parenthesized trinomial is a perfect square, $\left(x + 4\right)^2$ , and the loose constants combine to $-16 + 3 = -13$ :

x^2 + 8x + 3 = (x + 4)^2 - 13.

A quick check expands it back: $(x + 4)^2 - 13 = x^2 + 8x + 16 - 13 = x^2 + 8x + 3$ , the original. The constant dropped from $3$ to $-13$ precisely because $16$ was borrowed to build the square.

Check your understanding

Rewrite $x^2 - 10x + 18$ in completed-square form.

(x - 5)^2 - 7

(x - 5)^2 + 7

(x - 10)^2 - 82

(x - 5)^2 - 18

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:

x^2 + bx = -c.

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:

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

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. A positive number is the square of two different numbers, one positive and one negative, so both must be kept. That is where the $\pm$ sign comes from:

x + \frac{b}{2} = \pm\sqrt{\left(\frac{b}{2}\right)^2 - c}.

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

x = -\frac{b}{2} \pm \sqrt{\left(\frac{b}{2}\right)^2 - c}.

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.

Worked example 2 Solve $x^2 - 6x + 4 = 0$

This quadratic has no whole-number factor pair that works, so factoring is a dead end. Complete the square instead. Move the constant across:

x^2 - 6x = -4.

Here $b = -6$ , so $\tfrac{b}{2} = -3$ and $\left(\tfrac{b}{2}\right)^2 = 9$ . Add $9$ to both sides:

x^2 - 6x + 9 = -4 + 9 = 5.

The left side is the perfect square $(x - 3)^2$ , so take the square root of both sides, keeping both signs:

(x - 3)^2 = 5 \quad\Rightarrow\quad x - 3 = \pm\sqrt{5}.

Add $3$ to finish:

x = 3 \pm \sqrt{5}.

The two solutions are $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$ , both perfectly real even though neither is a whole number. This is the kind of answer a search for integer factors would never have found.

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.

Worked example 3 Solve $x^2 + 3x + 1 = 0$ (an odd middle coefficient)

Move the constant across, then halve the odd coefficient $b = 3$ :

x^2 + 3x = -1, \qquad \frac{b}{2} = \frac{3}{2}, \qquad \left(\frac{3}{2}\right)^2 = \frac{9}{4}.

Add $\tfrac{9}{4}$ to both sides. On the right, write $-1$ as $-\tfrac{4}{4}$ so the fractions combine:

x^2 + 3x + \frac{9}{4} = -1 + \frac{9}{4} = -\frac{4}{4} + \frac{9}{4} = \frac{5}{4}.

The left side is $\left(x + \tfrac{3}{2}\right)^2$ . Take the square root of both sides, remembering that $\sqrt{\tfrac{5}{4}} = \tfrac{\sqrt{5}}{2}$ since the denominator $4$ is a perfect square:

\left(x + \frac{3}{2}\right)^2 = \frac{5}{4} \quad\Rightarrow\quad x + \frac{3}{2} = \pm\frac{\sqrt{5}}{2}.

Subtract $\tfrac{3}{2}$ and combine over the common denominator $2$ :

x = -\frac{3}{2} \pm \frac{\sqrt{5}}{2} = \frac{-3 \pm \sqrt{5}}{2}.

The single fraction $\dfrac{-3 \pm \sqrt{5}}{2}$ holds both roots at once, which is the standard way to write them.

Check your understanding

Solve $x^2 + 4x + 1 = 0$ by completing the square.

x = -2 \pm \sqrt{3}

x = 2 \pm \sqrt{3}

x = -2 \pm \sqrt{5}

x = -4 \pm \sqrt{3}

Real, repeated, and complex roots

Every $b$ and $c$ you have used so far is an ordinary real number, and that stays true for the rest of this lesson. Every solve reaches the same crossroads: after isolating the square you have

\left(x + \frac{b}{2}\right)^2 = R, \qquad \text{where } R = \left(\frac{b}{2}\right)^2 - c.

The number $R$ on the right decides everything, because you are about to take its square root, and there are exactly three cases. (The next lesson gives this same quantity, scaled up, its own name, the discriminant; here it is just “the number on the right.”)

If $R > 0$ , 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 $R = 0$ , 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 $R < 0$ , 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 $x = -\tfrac{b}{2} \pm (\text{something})\,i$ .

Worked example 4 Solve $x^2 + 2x + 5 = 0$ (complex roots)

Move the constant across and add $\left(\tfrac{2}{2}\right)^2 = 1$ to both sides:

x^2 + 2x = -5, \qquad x^2 + 2x + 1 = -5 + 1 = -4.

The left side is $(x + 1)^2$ , and the right side is negative, which is the signal that the roots are complex:

(x + 1)^2 = -4 \quad\Rightarrow\quad x + 1 = \pm\sqrt{-4}.

Rewrite the square root of the negative using $i$ , as you learned two lessons ago: $\sqrt{-4} = i\sqrt{4} = 2i$ . Then

x + 1 = \pm 2i \quad\Rightarrow\quad x = -1 \pm 2i.

The solutions are the conjugate pair $x = -1 + 2i$ and $x = -1 - 2i$ . A quick check confirms the first: $(-1 + 2i)^2 + 2(-1 + 2i) + 5 = (1 - 4i + 4i^2) + (-2 + 4i) + 5 = (1 - 4i - 4) - 2 + 4i + 5 = 0$ .

This is the payoff the chapter was built for. Back when you only had real numbers, a quadratic with $R < 0$ 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 $R < 0$ .

Check your understanding

Solve $x^2 - 2x + 10 = 0$ by completing the square.

x = 1 \pm 3i

x = -1 \pm 3i

x = 1 \pm 9i

x = 1 \pm 3

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.

Worked example 5 Solve $2x^2 + 8x - 10 = 0$ and $3x^2 + 6x + 9 = 0$

For $2x^2 + 8x - 10 = 0$ , divide every term by the leading coefficient $2$ to make the $x^2$ term stand alone:

\frac{2x^2}{2} + \frac{8x}{2} - \frac{10}{2} = 0 \quad\Rightarrow\quad x^2 + 4x - 5 = 0.

Now complete the square. Move the constant across and add $\left(\tfrac{4}{2}\right)^2 = 4$ to both sides:

x^2 + 4x + 4 = 5 + 4 = 9 \quad\Rightarrow\quad (x + 2)^2 = 9.

Take the root of both sides, $x + 2 = \pm 3$ , so $x = 1$ or $x = -5$ , two real solutions.

For $3x^2 + 6x + 9 = 0$ , divide through by $3$ first:

x^2 + 2x + 3 = 0.

Add $\left(\tfrac{2}{2}\right)^2 = 1$ to both sides after moving the constant over:

x^2 + 2x + 1 = -3 + 1 = -2 \quad\Rightarrow\quad (x + 1)^2 = -2.

The right side is negative, so the roots are complex. Since $\sqrt{-2} = i\sqrt{2}$ ,

x + 1 = \pm i\sqrt{2} \quad\Rightarrow\quad x = -1 \pm i\sqrt{2}.

Dividing by $a$ at the very start is what kept both problems inside the method you already know.

Check your understanding

Solve $2x^2 + 4x + 10 = 0$ by completing the square.

x = -1 \pm 2i

x = -2 \pm 4i

x = 1 \pm 2i

x = -1 \pm 2

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

Adding $\left(\tfrac{b}{2}\right)^2$ to only one side. When you are solving an equation, the completing constant must be added to both sides to keep the balance. Adding it only on the left quietly changes the equation and gives a wrong root.
Confusing the equation move with the expression move. When you are rewriting an expression (no equals sign), there is no other side to balance against, so you add and subtract the constant in the same line. Only an equation lets you add it to both sides.
Forgetting to divide by $a$ first. With a leading coefficient $a \ne 1$ , you cannot complete the square on $ax^2 + bx$ directly. Divide the whole equation by $a$ so the $x^2$ term stands alone, then proceed.
Dropping the $\pm$ . A positive number is the square of two numbers, one positive and one negative, so $x + \tfrac{b}{2} = \pm\sqrt{R}$ . Writing only the positive root throws away one of the two solutions.
Taking the square root of a negative as if it were real. If $R < 0$ , the roots are complex. Write $\sqrt{-4} = 2i$ , not $2$ and not “no solution.” The negative under the root is the whole reason the answer carries an $i$ .
Halving $b$ in the constant but not in the binomial (or the reverse). The completed square is $\left(x + \tfrac{b}{2}\right)^2$ , with the same $\tfrac{b}{2}$ inside the binomial and squared in the constant. Using $b$ in one place and $\tfrac{b}{2}$ in the other breaks the identity.

Key takeaways

To complete the square on $x^2 + bx$ , add $\left(\tfrac{b}{2}\right)^2$ , the square of half the linear coefficient, giving $x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2$ .
Any quadratic with leading coefficient $1$ rewrites in completed-square form $x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 + \left(c - \left(\tfrac{b}{2}\right)^2\right)$ , by adding and subtracting that constant.
To solve, move $c$ across, add $\left(\tfrac{b}{2}\right)^2$ to both sides, and write the left as a square. Then take the square root of both sides, keeping both signs, and isolate $x$ : $x = -\tfrac{b}{2} \pm \sqrt{\left(\tfrac{b}{2}\right)^2 - c}$ .
The quantity $R = \left(\tfrac{b}{2}\right)^2 - c$ sorts the roots: $R > 0$ gives two real roots, $R = 0$ gives one repeated root $x = -\tfrac{b}{2}$ , and $R < 0$ gives a complex conjugate pair. The next lesson scales this same quantity up and names it the discriminant.
Because $\sqrt{-1} = i$ , a negative $R$ still yields solutions, so every quadratic has roots, real or complex.
If the leading coefficient is $a \ne 1$ , divide the equation by $a$ first, then complete the square on the resulting quadratic.

Completing the Square

Learning goals

The constant that completes a square

Rewriting a quadratic in completed-square form

Solving a quadratic by completing the square

Real, repeated, and complex roots

When the leading coefficient is not 1

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice