Completing the square — rewriting quadratics — Mathematics

Before this

This leads to

A quadratic that will not factorise

Try to factorise $x^{2} + 6 x + 11$ . You need two numbers with product 11 and sum 6. The only whole-number pairs are 1 and 11, or −1 and −11. Neither pair adds to 6. So the method from the last two lectures fails.

But the expression still has a lot to tell you. Here is a question it can answer: what is the smallest value $x^{2} + 6 x + 11$ can ever take? Try some values of $x$ .

x	−6	−5	−4	−3	−2	−1	0
x² + 6x + 11	11	6	3	2	3	6	11

The values fall, reach 2 at $x = - 3$ , then rise again in the same pattern. It looks as if 2 is the smallest value. But a table only shows the values you tried. Could a fraction such as $x = - 3.1$ go lower? By the end of this lecture you will be able to prove it cannot, with one line of algebra. The tool is called completing the square.

The last lecture ended with a promise. It spotted squares that were already there, as in $x^{2} - 9$ . This lecture does something different. It builds a square that is not there yet.

What a squared bracket looks like

Start from the other direction. Expand a squared bracket, $(x + p)^{2}$ , where $p$ is any number:

(x + p)^{2} = x^{2} + 2 p x + p^{2}

Look at the pattern carefully. The middle coefficient is $2 p$ , which is twice the number in the bracket. And the last term is $p^{2}$ , the square of that number. For example:

(x + 3)^{2} (x - 4)^{2} (x + 5)^{2} = x^{2} + 6 x + 9 = x^{2} - 8 x + 16 = x^{2} + 10 x + 25

Now read it backwards. If you see $x^{2} + 6 x$ , the 6 must be $2 p$ . So $p$ is half of 6, which is 3. That is the whole secret of the method. The number in the bracket is half the coefficient of $x$ .

But $x^{2} + 6 x + 11$ is not exactly $(x + 3)^{2}$ . The square gives $+ 9$ at the end, and we have $+ 11$ . So the expression is the square, plus 2 more:

x^{2} + 6 x + 11 = (x^{2} + 6 x + 9) + 2 = (x + 3)^{2} + 2

Expand it to check: $(x + 3)^{2} + 2 = x^{2} + 6 x + 9 + 2 = x^{2} + 6 x + 11$ . Yes.

Why it is called completing the square

The name is a picture. Draw $x^{2}$ as a square of side $x$ . Draw $6 x$ as two strips, each 3 by $x$ , and put one on the right and one underneath. The shape is almost a square of side $x + 3$ . Only one corner is missing, and that corner is 3 by 3.

Left: x² + 6x as a square and two strips, each 3 by x. Right: add a 3 by 3 corner and the shape becomes a square of side x + 3. So x² + 6x + 9 = (x + 3)², and x² + 6x = (x + 3)² − 9.

Adding the 9 "completes" the square. So $x^{2} + 6 x$ on its own is the big square without its corner:

x^{2} + 6 x = (x + 3)^{2} - 9

This picture is over a thousand years old. Around the year 820, the Baghdad scholar al-Khwārizmī wrote a book on solving equations. He had no letters for unknowns, so he wrote everything in words. One problem was: a square and ten of its roots make 39. In our notation, that is $x^{2} + 10 x = 39$ . He halved the 10 to get 5, and squared it to get 25. Then he added 25 to 39 to make 64, a perfect square of side 8. So $x + 5 = 8$ , and $x = 3$ . He justified each step with a drawing like the one above. The word "algebra" comes from al-jabr, a word in the title of his book.

Check his answer: $3^{2} + 10 \times 3 = 9 + 30 = 39$ . It works.

The general method

The same reasoning works for any quadratic $x^{2} + b x + c$ . Half of $b$ goes in the bracket. Squaring the bracket adds an extra $(\frac{b}{2})^{2}$ , so take that away again. Then the original $c$ stays.

Completing the square (when the x² coefficient is 1)

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

In words: halve the coefficient of $x$ and put it in the bracket. Subtract the square of that half. Add the original constant. The result is in the form $(x + p)^{2} + q$ , which is called completed square form.

Stop and think

A student writes $x^{2} + 6 x + 11 = (x + 3)^{2} + 11$ . Is that right?

Show the answer

No. The bracket $(x + 3)^{2}$ already contains a $+ 9$ , so the student has counted 9 extra. Test it with $x = 0$ . The left side is 11. The right side is $9 + 11 = 20$ . The correct form takes the 9 away: $(x + 3)^{2} - 9 + 11 = (x + 3)^{2} + 2$ . Testing one easy value, such as $x = 0$ , catches this mistake every time.

The proof that 2 is the smallest value

Now return to the opening question. We have $x^{2} + 6 x + 11 = (x + 3)^{2} + 2$ . Think about the two parts separately.

The part $(x + 3)^{2}$ is a square. A square of a real number is never negative. It is zero when the bracket is zero, which is when $x = - 3$ , and it is positive for every other $x$ . So $(x + 3)^{2} + 2$ is always at least 2. It equals exactly 2 at $x = - 3$ . Any other value of $x$ , including $- 3.1$ , gives something bigger.

That is the proof the table could not give. Completed square form shows the smallest value without any trial and error. In general:

(x + p)^{2} + q \geq q

The two sides are equal only when $x = - p$ . So the smallest value is $q$ , and it happens when $x = - p$ .

Notice the sign. The bracket $(x + 3)$ is zero at $x = - 3$ , not at $x = 3$ . On a graph of $y = x^{2} + 6 x + 11$ , the lowest point is $(- 3, 2)$ . That point is called the vertex of the curve, and the graphs unit uses it constantly. Move the sliders below to see the vertex move as the quadratic changes.

Explore — completed square form and the lowest point

Choose b and c in y = x² + bx + c. Watch the completed square form and the lowest point of the curve. Try an odd value of b, and try b = 0.

b: 6

c: 11

Four quadratics in completed square form, with their lowest points.

Quadratic	Completed square form	Smallest value
x² + 6x + 11	(x + 3)² + 2	2, at x = −3
x² − 8x + 7	(x − 4)² − 9	−9, at x = 4
x² + 5x + 1	(x + 5/2)² − 21/4	−21/4, at x = −5/2
x² − 6	x² − 6 (b = 0)	−6, at x = 0

Worked examples

Worked Example 1 — A negative coefficient of x

Write $x^{2} - 8 x + 7$ in completed square form. State its smallest value.

Half of $- 8$ is $- 4$ , so the bracket is $(x - 4)^{2}$ . It expands to $x^{2} - 8 x + 16$ , which has 16 at the end. Take the 16 away and add the 7:

x^{2} - 8 x + 7 = (x - 4)^{2} - 16 + 7 = (x - 4)^{2} - 9

Check with $x = 0$ : the original gives 7, and $16 - 9 = 7$ . The smallest value is $- 9$ , when $x = 4$ . Keep the sign with the half: the bracket is $x - 4$ , and it is zero at $x = + 4$ .

Worked Example 2 — An odd coefficient of x

Write $x^{2} + 5 x + 1$ in completed square form.

Half of 5 is $\frac{5}{2}$ . Do not round it to a decimal. Fractions keep the answer exact. The square of $\frac{5}{2}$ is $\frac{25}{4}$ .

x^{2} + 5 x + 1 = (x + \frac{5}{2})^{2} - \frac{25}{4} + 1 = (x + \frac{5}{2})^{2} - \frac{21}{4}

Here $1 = \frac{4}{4}$ , so $- \frac{25}{4} + \frac{4}{4} = - \frac{21}{4}$ . Check with $x = 0$ : $\frac{25}{4} - \frac{21}{4} = \frac{4}{4} = 1$ . The smallest value is $- \frac{21}{4}$ , which is $- 5.25$ , when $x = - 2.5$ .

Worked Example 3 — When the x² coefficient is not 1

Write $2 x^{2} + 12 x + 5$ in completed square form.

The halving rule only works when $x^{2}$ stands alone. So first take the 2 out of the first two terms only. Leave the constant outside:

2 x^{2} + 12 x + 5 = 2 (x^{2} + 6 x) + 5

Now complete the square inside the bracket. We already know $x^{2} + 6 x = (x + 3)^{2} - 9$ . Use square brackets to keep the whole thing together:

2 (x^{2} + 6 x) + 5 = 2 [(x + 3)^{2} - 9] + 5 = 2 (x + 3)^{2} - 18 + 5 = 2 (x + 3)^{2} - 13

The 2 multiplies everything in the square bracket, so the $- 9$ becomes $- 18$ . This is the step where most mistakes happen. Check by expanding: $2 (x^{2} + 6 x + 9) - 13 = 2 x^{2} + 12 x + 18 - 13 = 2 x^{2} + 12 x + 5$ . The smallest value is $- 13$ , when $x = - 3$ .

Worked Example 4 — A negative x² coefficient gives a largest value

Write $- x^{2} + 4 x + 1$ in completed square form. Does it have a smallest or a largest value?

Take out $- 1$ from the first two terms. Watch the sign of $4 x$ : $- x^{2} + 4 x = - (x^{2} - 4 x)$ .

- x^{2} + 4 x + 1 = - [(x - 2)^{2} - 4] + 1 = - (x - 2)^{2} + 4 + 1 = - (x - 2)^{2} + 5

Now the square is subtracted. $(x - 2)^{2}$ is never negative, so $- (x - 2)^{2}$ is never positive. The expression is at most 5, and it equals 5 when $x = 2$ . So it has a largest value of 5, and no smallest value. Check with $x = 0$ : $- 4 + 5 = 1$ , which matches the original.

Worked Example 5 — Completing the square, then factorising

Factorise $x^{2} + 2 x - 8$ using completed square form.

Half of 2 is 1, so $x^{2} + 2 x - 8 = (x + 1)^{2} - 1 - 8 = (x + 1)^{2} - 9$ .

This is the shape from Worked Example 5 of the last lecture: a squared bracket minus a square number. It is a difference of two squares, with $a = x + 1$ and $b = 3$ :

(x + 1)^{2} - 3^{2} = (x + 1 + 3) (x + 1 - 3) = (x + 4) (x - 2)

The product-and-sum method gives the same answer: product −8 and sum 2 give 4 and −2. So completing the square is not a separate topic. It joins up everything in this unit.

Here is where that joining up becomes powerful. Go back to an expression that does not factorise with whole numbers, such as $x^{2} + 4 x - 1$ . Half of 4 is 2, so:

x^{2} + 4 x - 1 = (x + 2)^{2} - 4 - 1 = (x + 2)^{2} - 5

Five is not a square number, but it is the square of $5$ . The last lecture said that whether you "can" factorise depends on which numbers you allow. Allow surds, and the difference of two squares still works:

(x + 2)^{2} - 5 = (x + 2 + 5) (x + 2 - 5)

Each bracket is zero for one value of $x$ . So the expression is zero when $x = - 2 - 5$ or $x = - 2 + 5$ , which are about −4.24 and 0.24. That is a quadratic equation solved exactly, without a formula. The lecture on solving quadratics by completing the square takes this idea and makes it a full method.

Stop and think

Can $x^{2} - 8 x + 7$ ever equal $- 10$ ?

Show the answer

No. Worked Example 1 showed $x^{2} - 8 x + 7 = (x - 4)^{2} - 9$ . The square is never negative, so the expression is never less than $- 9$ . And $- 10$ is less than $- 9$ . Completed square form answers this at once, with no trial and error.

Now look at a student's working for a quadratic whose $x^{2}$ coefficient is 3.

Spot the error — completing the square on 3x² + 12x + 2

One line of this working is wrong. Tap the line you think is wrong. Every line has an explanation.

The working, with the mistake marked and corrected:

Line 1. 3x² + 12x + 2 = 3(x² + 4x) + 2. (Correct.)
Line 2. Half of 4 is 2, and (x + 2)² = x² + 4x + 4. (Correct.)
Line 3. So 3(x² + 4x) + 2 = 3[(x + 2)² − 4] + 2. (Correct.)
Line 4. = 3(x + 2)² − 4 + 2. The mistake: the 3 multiplies the −4 as well, so it should be −12.
Line 5. = 3(x + 2)² − 2. (This follows from the mistake. The correct answer is 3(x + 2)² − 12 + 2 = 3(x + 2)² − 10.)

Three classic mistakes

Trap 1 — Forgetting to take the square away

$x^{2} + 6 x + 11$ is not $(x + 3)^{2} + 11$ . The bracket brings its own $+ 9$ , so you must subtract it: $(x + 3)^{2} - 9 + 11 = (x + 3)^{2} + 2$ . Test with $x = 0$ to catch this.

Trap 2 — Not multiplying by the x² coefficient

In $2 x^{2} + 12 x + 5 = 2 [(x + 3)^{2} - 9] + 5$ , the 2 multiplies the $- 9$ too. The answer is $2 (x + 3)^{2} - 13$ , not $2 (x + 3)^{2} - 4$ . Also, never halve the 12 before taking out the 2. The halving rule needs $x^{2}$ on its own.

Trap 3 — The wrong sign for the lowest point

$(x + 3)^{2} + 2$ has its smallest value at $x = - 3$ , not $x = 3$ . Ask: what value of $x$ makes the bracket zero? For $(x - 4)^{2} - 9$ , it is $x = 4$ .

Try it yourself

Work each one on paper first. Check each answer by expanding, or with $x = 0$ .

Write $x^{2} + 10 x + 30$ in completed square form. What is its smallest value, and for which $x$ ?

Hint 1
Half of 10 is 5. What does $(x + 5)^{2}$ expand to?

Hint 2
$(x + 5)^{2} = x^{2} + 10 x + 25$ . You need 30 at the end, not 25.

Full answer
$x^{2} + 10 x + 30 = (x + 5)^{2} - 25 + 30 = (x + 5)^{2} + 5$ . The smallest value is 5, when $x = - 5$ . Check with $x = 0$ : $25 + 5 = 30$ .
Write $x^{2} - 3 x + 1$ in completed square form. Give exact fractions.

Hint 1
Half of $- 3$ is $- \frac{3}{2}$ .

Hint 2
$(- \frac{3}{2})^{2} = \frac{9}{4}$ , and $1 = \frac{4}{4}$ .

Full answer
$x^{2} - 3 x + 1 = (x - \frac{3}{2})^{2} - \frac{9}{4} + \frac{4}{4} = (x - \frac{3}{2})^{2} - \frac{5}{4}$ . The smallest value is $- \frac{5}{4}$ , when $x = \frac{3}{2}$ . Check with $x = 0$ : $\frac{9}{4} - \frac{5}{4} = 1$ .
Write $3 x^{2} - 12 x + 7$ in completed square form.

Hint 1
Take 3 out of the first two terms: $3 (x^{2} - 4 x) + 7$ .

Hint 2
$x^{2} - 4 x = (x - 2)^{2} - 4$ . Remember that the 3 multiplies the $- 4$ .

Full answer
$3 [(x - 2)^{2} - 4] + 7 = 3 (x - 2)^{2} - 12 + 7 = 3 (x - 2)^{2} - 5$ . Check by expanding: $3 (x^{2} - 4 x + 4) - 5 = 3 x^{2} - 12 x + 12 - 5 = 3 x^{2} - 12 x + 7$ . The smallest value is $- 5$ , when $x = 2$ .
A farmer has 20 m of fencing. She makes a rectangular pen against a long wall, so the fence forms only three sides. The two sides that meet the wall are each $x$ m long. Show that the area is $20 x - 2 x^{2}$ square metres. Then use completed square form to find the largest possible area.

Hint 1
The two short sides use $2 x$ m of fence, so the long side is $20 - 2 x$ m. Area = length × width.

Hint 2
Write $- 2 x^{2} + 20 x = - 2 (x^{2} - 10 x)$ , then complete the square inside the bracket.

Full answer
Area $= x (20 - 2 x) = 20 x - 2 x^{2}$ . Then $- 2 (x^{2} - 10 x) = - 2 [(x - 5)^{2} - 25] = - 2 (x - 5)^{2} + 50$ . The term $- 2 (x - 5)^{2}$ is never positive, so the area is at most 50 m². It equals 50 m² when $x = 5$ . The pen is then 5 m by 10 m. Check: $5 \times 10 = 50$ , and $5 + 10 + 5 = 20$ m of fence.

The bigger idea

Completing the square turns a quadratic into "a square plus a number". That is the most revealing shape a quadratic can take. A square is never negative, so the smallest or largest value can be read off at once. Set the expression equal to zero and you can solve it, even when it does not factorise. Do the same steps with letters instead of numbers and you get the quadratic formula. The graphs unit uses completed square form to find the vertex of any parabola in one line.

This completes the unit on expanding and factorising. The next unit, on quadratic equations, uses every method from this unit. Completing the square will be at the centre of it.

Key concepts in this lesson

$(x + p)^{2} = x^{2} + 2 p x + p^{2}$ , so the number in the bracket is half the coefficient of $x$ .
$x^{2} + b x + c = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c$ : halve, square, take the square away, add the constant.
The name comes from a picture: $x^{2} + b x$ is a square with one corner missing, and the corner has area $(\frac{b}{2})^{2}$ .
$(x + p)^{2} + q$ has smallest value $q$ , when $x = - p$ . With a negative $x^{2}$ coefficient, the value $q$ is the largest instead.
When the $x^{2}$ coefficient is not 1, take it out of the first two terms first, and multiply the subtracted square by it.