Use fractions: half of 5 is 52, and ( 52 )^2 = 254.

1MA1 · A11, A18AlgebraHigherGrades 7–9

Completing the Square

Q: Why is it called completing the square?

Geometrically, x^2 + bx is a square with two rectangles attached; adding ( b2 )^2 fills in the missing corner to make a complete square.

Q: Is completing the square better than the formula?

For solving, both give the same answers. Completing the square also gives the turning point, and the quadratic formula is actually derived from it.

Every quadratic $x^{2} + bx + c$ can be rewritten in the form $(x + p)^{2} + q$ . That rearrangement — $\textbf{completing the square}$ — does two jobs at once: it solves the equation exactly, and it tells you the turning point of the graph without plotting a single point.

50 min Quadratics

The big picture

The idea comes from expanding a squared bracket: $(x + 3)^{2} = x^{2} + 6x + 9$ . The number inside the bracket is always half the coefficient of $x$ . So to rewrite $x^{2} + 6x - 7$ , start with $(x + 3)^{2}$ , which gives $x^{2} + 6x + 9$ , and subtract the unwanted 9.

In completed-square form, the squared term can never be negative, so its smallest value is 0. That immediately gives the minimum of the quadratic and where it occurs — the turning point — which Edexcel links to graph sketching and, at A-Level, to circles.

What you'll learn

Write $x^{2} + bx + c$ in the form $(x + p)^{2} + q$
Complete the square when the $x^{2}$ coefficient is not 1
Solve quadratic equations by completing the square, giving exact answers
Find the turning point of a quadratic graph from completed-square form

The method for x² + bx + c

Halve the coefficient of $x$ to get $p$ , and write $(x + p)^{2}$ .

Expanding $(x + p)^{2}$ produces an extra $p^{2}$ , so subtract it, then add the original constant.

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

x^{2} + bx + c = \left(x + \tfrac{b}{2}\right)^{2} - \tfrac{b^{2}}{4} + c

Completing the square.

Example 1Write

x^{2} + 6x - 7

in the form

(x + p)^{2} + q

1Half of 6 is 3:

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

2Subtract 9 and include

-7

(x + 3)^{2} - 9 - 7

Answer

(x + 3)^{2} - 16

Your turnSame method, new numbers — try it before moving on.

Write $x^2 + 8x + 3$ in the form $(x + p)^2 + q$ .

Your answer

Type it like (x+3)^2-5.

Show worked solution

Solution

1Half of 8 is 4: $(x + 4)^2 = x^2 + 8x + 16$ .
2 $16 - 13 = 3$ , so $(x + 4)^2 - 13$ .

Answer: $(x + 4)^2 - 13$

Example 2Write

x^{2} - 5x + 2

in completed-square form.

1Half of

-5

-\tfrac52

\left(x - \tfrac52\right)^{2} = x^{2} - 5x + \tfrac{25}{4}

2Subtract

\tfrac{25}{4}

and add 2.

Answer

\left(x - \tfrac52\right)^{2} - \tfrac{17}{4}

Tip — Check by expanding your answer. It takes seconds and catches sign errors in $q$ .

When the x² coefficient is not 1

Factor the coefficient out of the $x^{2}$ and $x$ terms only, complete the square inside the bracket, then multiply back out.

For $2x^{2} - 8x + 5$ : $2(x^{2} - 4x) + 5 = 2\big[(x - 2)^{2} - 4\big] + 5 = 2(x - 2)^{2} - 8 + 5$ .

Example 1Write

2x^{2} - 8x + 5

in the form

a(x + p)^{2} + q

1Factor 2 from the first two terms:

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

2Complete the square inside:

x^{2} - 4x = (x - 2)^{2} - 4

2(x - 2)^{2} - 8 + 5

Answer

2(x - 2)^{2} - 3

Your turnSame method, new numbers — try it before moving on.

Write $2x^2 + 12x + 7$ in the form $a(x + p)^2 + q$ .

Your answer

Type it like 2(x+3)^2-11.

Show worked solution

Solution

1 $2(x^2 + 6x) + 7 = 2[(x + 3)^2 - 9] + 7$ .
2 $= 2(x + 3)^2 - 18 + 7 = 2(x + 3)^2 - 11$ .

Answer: $2(x + 3)^2 - 11$

Why this works

The $-4$ inside the square bracket is multiplied by 2 when you expand the outer bracket, giving $-8$ . Forgetting that multiplication is the most common error with non-unit coefficients.

Solving by completing the square

Write the quadratic in completed-square form, set it equal to zero, and isolate the squared bracket.

Square root both sides, remembering $\pm$ , then solve.

This gives exact answers in surd form, which is why questions ask for it by name.

Example 1Solve

x^{2} - 4x - 1 = 0

by completing the square, giving exact answers.

(x - 2)^{2} - 4 - 1 = 0

, so

(x - 2)^{2} = 5

x - 2 = \pm\sqrt5

Answer

x = 2 \pm \sqrt5

Your turnSame method, new numbers — try it before moving on.

Solve $x^2 + 6x + 2 = 0$ by completing the square. Give exact answers.

Your answer

Type both, e.g. -3+sqrt7, -3-sqrt7.

Show worked solution

Solution

1 $(x + 3)^2 - 9 + 2 = 0$ , so $(x + 3)^2 = 7$ .
2 $x + 3 = \pm\sqrt{7}$ , so $x = -3 \pm \sqrt{7}$ .

Answer: $x = -3 \pm \sqrt{7}$

Tip — Leave answers in surd form unless told to round. "Exact" means $2 + \sqrt5$ , not $4.24$ .

Turning points

In $(x + p)^{2} + q$ , the squared term is at least 0 and equals 0 when $x = -p$ . So the minimum value is $q$ , at $x = -p$ .

The turning point of $y = (x + p)^{2} + q$ is $(-p, q)$ .

For $y = a(x + p)^{2} + q$ , the turning point is still $(-p, q)$ ; it is a minimum if $a > 0$ and a maximum if $a < 0$ .

Example 1Find the turning point of

y = x^{2} + 6x - 7

and state whether it is a maximum or minimum.

1From before,

y = (x + 3)^{2} - 16

2Minimum when

x + 3 = 0

AnswerMinimum at

(-3, -16)

Your turnSame method, new numbers — try it before moving on.

Find the turning point of $y = x^2 - 10x + 18$ .

Your answer

Type coordinates like (2, -3).

Show worked solution

Solution

1 $y = (x - 5)^2 - 25 + 18 = (x - 5)^2 - 7$ .
2Minimum at $(5, -7)$ .

Answer: $(5, -7)$

Think like an examiner

Show the intermediate form $(x + p)^{2} - p^{2} + c$ before simplifying.
Include $\pm$ when square rooting to solve.
Give turning points as coordinates and state the nature if asked.

Completed square

(x + p)^{2} + q

Completed-square form.

(-p,\ q)

Turning point.

p = \tfrac{b}{2}

Half the x coefficient.

Watch out for these

Writing

x^{2} + 6x - 7 = (x + 6)^{2} - 43

Halve the coefficient:

(x + 3)^{2} - 16

Adding

p^{2}

instead of subtracting it.

The bracket already creates

+p^{2}

, so subtract it.

Giving the turning point of

(x - 2)^{2} + 5

(-2, 5)

The

x

-coordinate has the opposite sign:

(2, 5)

Solving

(x - 2)^{2} = 5

x = 2 + \sqrt5

only.

Use

\pm

x = 2 \pm \sqrt5

Stretch yourself

The curve $y = x^{2} + bx + c$ has a turning point at $(3, -5)$ . Find $b$ and $c$ , and state the $y$ -intercept.

Hint — Write the curve in completed-square form using the turning point, then expand.

Reveal solution

SolutionA turning point at

(3, -5)

means

y = (x - 3)^{2} - 5

. Expanding:

x^{2} - 6x + 9 - 5 = x^{2} - 6x + 4

. So

b = -6

and

c = 4

, and the

y

-intercept is

(0, 4)

. Check: at

x = 3

9 - 18 + 4 = -5

✓.

\blacksquare

Questions students ask

Why is it called completing the square?

Geometrically,

x^{2} + bx

is a square with two rectangles attached; adding

\left(\tfrac{b}{2}\right)^{2}

fills in the missing corner to make a complete square.

Is completing the square better than the formula?

For solving, both give the same answers. Completing the square also gives the turning point, and the quadratic formula is actually derived from it.

What if

b

is odd?

Use fractions: half of 5 is

\tfrac52

, and

\left(\tfrac52\right)^{2} = \tfrac{25}{4}

Key takeaways

$x^{2} + bx + c = \left(x + \tfrac{b}{2}\right)^{2} - \left(\tfrac{b}{2}\right)^{2} + c$ .
For $ax^{2} + bx + c$ , factor $a$ from the first two terms first.
Solve by isolating the square, square rooting with $\pm$ .
Turning point of $(x + p)^{2} + q$ is $(-p, q)$ .
Positive $a$ gives a minimum; negative $a$ gives a maximum.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 73 marks

(a)Write $x^2 - 12x + 40$ in the form $(x - a)^2 + b$ .[2]
Show mark scheme (a)
- M1 $(x - 6)^2$
- A1 $(x - 6)^2 + 4$
Answer: $(x - 6)^2 + 4$
(b)Hence explain why $x^2 - 12x + 40 > 0$ for all $x$ .[1]
Show mark scheme (b)
- B1 $(x - 6)^2 \ge 0$ , so the expression is at least 4

Question 2Grade 83 marks

Write $3x^2 - 12x + 5$ in the form $a(x + b)^2 + c$ .[3]
Show mark scheme
- M1 $3(x^2 - 4x) + 5$
- M1 $3[(x - 2)^2 - 4] + 5$
- A1 $3(x - 2)^2 - 7$
Answer: $3(x - 2)^2 - 7$

Question 3Grade 93 marks

Solve $x^2 + 4x - 3 = 0$ by completing the square. Give your answers in the form $a \pm \sqrt{b}$ .[3]
Show mark scheme
- M1 $(x + 2)^2 - 7 = 0$
- M1 $x + 2 = \pm\sqrt{7}$
- A1 $x = -2 \pm \sqrt{7}$
Answer: $x = -2 \pm \sqrt{7}$

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