Learning to Complete the Square

What this move does

The puzzle above says just one thing:

x^{2} + b x + (\frac{b}{2})^{2} = (x + \frac{b}{2})^{2}

$x^{2} + b x$ is a square with a corner missing (an L-shape). The missing corner is a small square of side $\frac{b}{2}$ : half the coefficient of the linear term. Fill it in and you get a perfect square. This move is called completing the square:

Look at the coefficient $b$ of the linear term and take half of it, $\frac{b}{2}$ .
Add $(\frac{b}{2})^{2}$ and subtract it again at the same time (or add it to both sides of an equation).
Write the first three terms as $(x + \frac{b}{2})^{2}$ and combine the constants that are left.

When $b$ is negative the picture cannot show a “negative length”, but the algebra is exactly the same: $x^{2} - 6 x = (x - 3)^{2} - 9$ .

Predict first

$x^{2} + 10 x$ is still missing a corner. How much must you add to make it a perfect square?

5

10

25

100

Choose an answer first; then you can see the result.

Work through one

The most direct use of completing the square is solving equations. The equation $x^{2} + 6 x = 16$ says: the area of the L-shape is $16$ .

Example: solve x² + 6x = 16

Step 1 / 5

Try any x first
In the picture $x = 4$ and the area is $4^{2} + 6 \times 4 = 40$ , much bigger than $16$ . Trying values one by one is too slow, so change tactics: complete the L-shape into a square.

$x^{2} + 6 x = 16$
Split the linear term in half
Move one strip underneath
Add 9 to both sides
Take square roots

Check $x = - 8$ : $(- 8)^{2} + 6 \times (- 8) = 64 - 48 = 16$ . Correct. The picture can only show the positive root; the algebra finds both.

When the leading coefficient is not 1

Take the leading coefficient out first, and complete the square inside the bracket.

Example: complete the square for 2x² − 12x + 7

Step 1 / 4

Take out the 2
Taking it out of the first two terms is enough; leave the constant outside for now.

$2 x^{2} - 12 x + 7 = 2 (x^{2} - 6 x) + 7$
Fill the corner inside the bracket
Write it as a square
Multiply out the bracket

The floor of squares

Completing the square has a second use, and it is even neater. The square of any real number is never less than 0: $0$ is a floor you cannot break through.

So if several squares add up to $0$ , there is only one way it can happen: every one of them is 0. Drag the sliders below and see whether you can make one bar “dig underground” to cancel another.

Make the “Total” equal

0

. How many ways are there?

So one equation can pin down two unknowns at once, as long as you first rewrite it as a sum of squares:

Example: if x² + y² − 6x + 4y + 13 = 0, find x + y

Step 1 / 6

One equation, two unknowns?
Usually an equation like this has infinitely many solutions. But do not give up yet: group the $x$ terms together and the $y$ terms together.

$(x^{2} - 6 x) + (y^{2} + 4 y) + 13 = 0$
How much does each need?
Write it as a sum of squares
Push down the first bar
Push down the second bar
Answer the question

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: the equation “sum of squares $= 0$ ” has exactly one solution

$x^{2} + y^{2} - 2 x + 4 y + 5 = 0$ : the constant $5 = 1 + 4$ is shared out exactly between the two squares. Push both bars down to the floor and the total is $0$ ; the only solution is $x = 1$ , $y = - 2$ .

Practice

Exercise 1Basic

$x^{2} - 8 x + m$ is a perfect square. Find $m$ .

Exercise 2Basic

Solve $x^{2} + 4 x - 12 = 0$ by completing the square. (Separate the two roots with a comma.)

Exercise 3Advanced

Given $a^{2} + b^{2} - 4 a + 10 b + 29 = 0$ , find $a + b$ .

Exercise 4Challenge

$x$ and $y$ can be any real numbers. Find the minimum value of $x^{2} - 2 x y + 2 y^{2} - 4 y + 5$ .

What this move does

Work through one

Example: solve x² + 6x = 16

When the leading coefficient is not 1

Example: complete the square for 2x² − 12x + 7

The floor of squares

Example: if x² + y² − 6x + 4y + 13 = 0, find x + y

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Adding without subtracting

Forgetting the 2 outside the bracket

“One equation, two unknowns, so it cannot be solved”