Completing the square and the vertex — Algebra and functions · AQA 7357

Section 1 of 14

Learning objectives

Write $x^2+bx+c$ in the form $(x+p)^2+q$ .
Write $ax^2+bx+c$ with $a\ne1$ in the form $a(x+p)^2+q$ , including when $a$ is negative.
Read the vertex, the line of symmetry and the least or greatest value from the completed square.
Sketch a parabola, showing its intercepts and vertex.
Find unknown constants from information about the vertex, and solve "maximum" problems in context.

Pause and think

Warm-up: expand $(x+5)^2$ . Which number in your answer is "extra" compared with $x^2+10x$ ?

Section 2 of 14

Core idea

Start hereHow high does the ball go?

You kick a ball straight up. Its height in metres after $t$ seconds is $h=1+20t-5t^2$ . Two questions: how high does it get, and when? Plugging in numbers ( $t=1$ gives $16$ , $t=2$ gives $21$ , $t=3$ gives $16$ ) suggests the top is near $t=2$ , but guessing never proves anything. What if the top were at $t=2.01$ ? This lesson gives you a way to rewrite the formula so the highest point simply appears in the expression, with no guessing, graph or calculator. By the end you'll show the answer is exactly $21$ m at exactly $t=2$ s.

The big idea

Any quadratic $ax^2+bx+c$ can be rewritten as $a(x+p)^2+q$ . This is called completing the square. In this form $x$ appears only once, inside a square. Because a square can never be negative, you can read off the turning point, the line of symmetry and the smallest (or largest) value straight away. You can also solve the equation with a single square root.

1. The language: what every symbol means

A quadratic expression in $x$ is anything that can be written as

General quadratic

ax^2+bx+c,\qquad a\neq0

$x$: the variable: the number that can change (time, length, anything).
$a$: the leading coefficient: the number multiplying $x^2$ . It must not be $0$ , or there'd be no $x^2$ and the graph would be a straight line.
$b$: the coefficient of $x$ (it may be $0$ ).
$c$: the constant term: the number with no $x$ (it may be $0$ ).

$x^2+6x+5$ : $a=1$ , $b=6$ , $c=5$ .
$2x^2-3x$ : $a=2$ , $b=-3$ , $c=0$ . (The sign belongs to the coefficient.)
$7-x^2$ : $a=-1$ , $b=0$ , $c=7$ . Order doesn't matter; read off the number in front of each power.

The graph of $y=ax^2+bx+c$ is a parabola: a symmetrical curve that opens up (∪) when $a>0$ and down (∩) when $a<0$ . Its turning point is the vertex, and the vertical line through it is the line of symmetry.

The sign of a decides the shape

a = -2.0

Positive

a

: a ∪ with a lowest point. Negative

a

: a ∩ with a highest point. The bigger

|a|

is, the narrower the curve.

Key takeawayThe central problem In

ax^2+bx+c

the variable appears twice (in

x^2

and in

x

), and the two terms pull in different directions. That's why you can't just "undo" it to find

x

, or see where the vertex is. Our goal: rewrite it so $x$ appears only once.

Pause and think

Write down $a$ , $b$ and $c$ for $3-5x+2x^2$ , and say whether its graph is ∪ or ∩.

2. Perfect squares: the building block

A perfect square is a bracket multiplied by itself, such as $(x+3)^2$ . It contains $x$ only once, which is exactly the form we want. Let's expand one slowly to see what it looks like when multiplied out.

Expanding

(x+3)^2

$(x+3)^2$
$=(x+3)(x+3)$
Squaring means multiplying by itself.
$=x\cdot x+x\cdot3+3\cdot x+3\cdot3$
Multiply every term in the first bracket by every term in the second (four products).
$=x^2+3x+3x+9$
Work out each product.
$=x^2+6x+9$
Collect the like terms $3x+3x=6x$ .

The middle coefficient

6

is double

3

; the constant

9

is

3

squared.

Is that a coincidence? Replace $3$ by any number $p$ and repeat the same steps:

Proof of the perfect-square pattern

$(x+p)^2=(x+p)(x+p)$
Definition of squaring.
$=x^2+px+px+p^2$
Four products: $x\cdot x$ , $x\cdot p$ , $p\cdot x$ , $p\cdot p$ .
$=x^2+2px+p^2$
$px+px=2px$ .

For every number

p

:

\;(x+p)^2=x^2+2px+p^2

. The

x

-coefficient is always

2p

and the constant is always

p^2

.

(x + 3)² as a picture: a square of side x + 3

The big square splits into four pieces:

x^2

, two strips of

3x

(that is where the "double" comes from) and a little

3\times3

corner, the

9

.

Common misconception $(x+3)^2$ is not $x^2+9$ Squaring a sum does not square each part separately. Check with

x=1

:

(1+3)^2=16

, but

1^2+9=10

. The missing

6

is the two

3x

strips in the picture.

Pause and think

Expand $(x-5)^2$ using the pattern with $p=-5$ .

3. Running the pattern backwards

Now reverse it. Suppose you're given $x^2+bx$ and want to write it using a perfect square. Matching $x^2+bx$ with $x^2+2px$ tells us $2p=b$ , so $p=\frac b2$ : the number in the bracket is half the coefficient of $x$ . But $(x+\frac b2)^2$ brings an extra $(\frac b2)^2$ with it, so we must take that away again.

The completing-the-square identity

$\left(x+\tfrac b2\right)^2=x^2+2\cdot\tfrac b2\cdot x+\left(\tfrac b2\right)^2$
The pattern from section 2 with $p=\frac b2$ .
$\left(x+\tfrac b2\right)^2=x^2+bx+\left(\tfrac b2\right)^2$
$2\cdot\frac b2=b$ .
$\left(x+\tfrac b2\right)^2-\left(\tfrac b2\right)^2=x^2+bx$
Subtract $\left(\frac b2\right)^2$ from both sides.

x^2+bx=\left(x+\dfrac b2\right)^2-\left(\dfrac b2\right)^2

. Memory hook: halve it, square it, take it away.

In the picture, $x^2+bx$ is the big square missing its corner. Completing the square means adding the corner to make a full square, then subtracting it so the value hasn't changed. That's where the name comes from.

Worked exampleComplete the square on $x^2+6x+5$

Write $x^2+6x+5$ in the form $(x+p)^2+q$ .

Goal: Find the numbers $p$ and $q$ .
Strategy: Apply "halve, square, take away" to the $x^2+6x$ part only, then add the $5$ back.

Identify the $x$ -coefficient: $b=6$ . Halve it: $\frac62=3$ .
The bracket must be $(x+\frac b2)$ .
$x^2+6x=(x+3)^2-3^2$
Square the $3$ and subtract it.
$x^2+6x+5=(x+3)^2-9+5$
Add $5$ to both sides.
$=(x+3)^2-4$
$-9+5=-4$ .

Answer

(x+3)^2-4

, so

p=3

,

q=-4

.

✓ CheckExpand:

(x+3)^2-4=x^2+6x+9-4=x^2+6x+5

✓. Also try

x=0

:

9-4=5

and

0+0+5=5

✓.

Worked exampleA negative and an odd coefficient: $x^2-5x+1$

Write $x^2-5x+1$ in the form $(x+p)^2+q$ .

Goal: Find $p$ and $q$ , keeping exact fractions.
Strategy: Same three moves. Halve $-5$ with its sign; don't convert to decimals.

$b=-5$ , so $\frac b2=-\frac52$ .
Keep the minus sign.
$x^2-5x=\left(x-\tfrac52\right)^2-\left(-\tfrac52\right)^2=\left(x-\tfrac52\right)^2-\tfrac{25}4$
Halve, square ( $(-\frac52)^2=+\frac{25}4$ , a square is positive), take away.
$x^2-5x+1=\left(x-\tfrac52\right)^2-\tfrac{25}4+\tfrac44$
Add $1$ , written as $\frac44$ so the fractions combine.
$=\left(x-\tfrac52\right)^2-\tfrac{21}4$
$-25+4=-21$ .

Answer

\left(x-\frac52\right)^2-\frac{21}4

.

✓ CheckExpand:

x^2-5x+\frac{25}4-\frac{21}4=x^2-5x+\frac44=x^2-5x+1

✓.

Common misconceptionForgetting to take it away Writing

x^2+6x+5=(x+3)^2+5

adds an extra

9

out of nowhere. The square you create must be paid back:

(x+3)^2-9+5

.

Pause and think

Complete the square: $x^2+12x-7$ .

4. Reading the vertex

Here's the payoff. Take $y=(x+3)^2-4$ . The only part that changes with $x$ is $(x+3)^2$ . The key fact: a real number squared is never negative.

Finding the minimum of

y=(x+3)^2-4

$(x+3)^2\ge0\quad\text{for every real }x$
Any real number squared is $\ge0$ .
$(x+3)^2-4\ge-4$
Subtract $4$ from both sides of the inequality.
$y\ge-4$
The left-hand side is $y$ .
$y=-4\iff(x+3)^2=0\iff x=-3$
Equality only when the square is exactly zero.

The least value of

y

is

-4

, reached only at

x=-3

. So the vertex is

(-3,-4)

and the line of symmetry is

x=-3

.

y = x² + 6x + 5 = (x + 3)² − 4

The roots

-5

and

-1

are each 2 units from the line of symmetry

x=-3

. The

y

-intercept

(0,5)

comes from putting

x=0

.

Vertex rule

y=a(x+p)^2+q\;\Longrightarrow\;\text{vertex }(-p,\,q),\quad\text{line of symmetry }x=-p

$-p$: the $x$ that makes the bracket zero. Note the opposite sign to the number in the bracket.
$q$: the minimum value if $a>0$ , or the maximum value if $a<0$ .

Sliding the parabola: y = (x − h)² + k

h = -3.0

Compared with

y=x^2

(dashed), the curve has slid

h

to the right and

2

down. Its shape never changes.

Common misconceptionThe sign flip

(x-4)^2+1

has its vertex at

(4,1)

, not

(-4,1)

. Ask: "what

x

makes the bracket zero?"

x-4=0

gives

x=4

. The

q

keeps its own sign.

Edge caseWhen a is negative If

a<0

, e.g.

y=-2(x-1)^2+5

, then

-2(x-1)^2\le0

, so

y\le5

: the vertex

(1,5)

is a maximum. Same reading, opposite meaning.

Pause and think

Find the vertex and line of symmetry of $y=x^2-10x+30$ .

5. When the coefficient of x² is not 1

$(x+p)^2$ always begins with $x^2$ , never $2x^2$ . So when $a\neq1$ , first take $a$ out as a factor of the $x^2$ and $x$ terms, complete the square inside the bracket, then multiply back.

Worked example $2x^2+12x+7$ in the form $a(x+p)^2+q$

Write $2x^2+12x+7$ as $a(x+p)^2+q$ and state the vertex of $y=2x^2+12x+7$ .

Goal: Find $a$ , $p$ , $q$ ; then the vertex $(-p,q)$ .
Strategy: Factor $2$ from the first two terms, complete the square inside, expand the outer bracket carefully.

$2x^2+12x+7=2\left(x^2+6x\right)+7$
Take out $2$ : $2x^2=2\cdot x^2$ and $12x=2\cdot6x$ . The $7$ stays outside.
$=2\left[(x+3)^2-9\right]+7$
Complete the square inside: halve $6$ , square, take away.
$=2(x+3)^2-18+7$
Multiply the $2$ into both parts of the square bracket: $2\times(-9)=-18$ .
$=2(x+3)^2-11$
$-18+7=-11$ .

Answer

2(x+3)^2-11

; vertex

(-3,-11)

, a minimum since

a=2>0

.

✓ CheckExpand:

2(x^2+6x+9)-11=2x^2+12x+18-11=2x^2+12x+7

✓.

Common misconceptionForgetting to multiply the −9 by a

2[(x+3)^2-9]+7

is not

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

. The

2

multiplies everything in the square bracket.

6. The general formula, and where the quadratic formula comes from

Doing the same steps with letters instead of numbers gives a rule that works for every quadratic, and proves the quadratic formula.

Completing the square on

ax^2+bx+c

$ax^2+bx+c=a\left(x^2+\tfrac bax\right)+c$
Factor $a$ out of the first two terms ( $a\neq0$ , so dividing by $a$ is allowed).
$=a\left[\left(x+\tfrac b{2a}\right)^2-\tfrac{b^2}{4a^2}\right]+c$
Halve $\frac ba$ to get $\frac b{2a}$ ; square it: $\frac{b^2}{4a^2}$ ; take it away.
$=a\left(x+\tfrac b{2a}\right)^2-\tfrac{b^2}{4a}+c$
Multiply $a$ in: $a\cdot\frac{b^2}{4a^2}=\frac{b^2}{4a}$ .
$=a\left(x+\tfrac b{2a}\right)^2+\tfrac{4ac-b^2}{4a}$
Common denominator: $c=\frac{4ac}{4a}$ .

Vertex at

x=-\dfrac b{2a}

, with

y

-value

\dfrac{4ac-b^2}{4a}

.

Now set the expression equal to zero and solve. Every step is legal algebra, so the result works for any quadratic equation.

Proof of the quadratic formula

$a\left(x+\tfrac b{2a}\right)^2-\tfrac{b^2-4ac}{4a}=0$
Start from the completed square (rewritten with $b^2-4ac$ on top), set equal to $0$ .
$a\left(x+\tfrac b{2a}\right)^2=\tfrac{b^2-4ac}{4a}$
Add $\frac{b^2-4ac}{4a}$ to both sides.
$\left(x+\tfrac b{2a}\right)^2=\tfrac{b^2-4ac}{4a^2}$
Divide both sides by $a$ .
$x+\tfrac b{2a}=\pm\tfrac{\sqrt{b^2-4ac}}{2a}$
Square root both sides; remember both $+$ and $-$ . ( $\sqrt{4a^2}=2|a|$ , and the $\pm$ covers the sign.)
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Subtract $\frac b{2a}$ from both sides and combine over $2a$ .

The quadratic formula is just completing the square done once and for all.

Edge caseWhen $b^2-4ac<0$ The step "square root both sides" needs

\frac{b^2-4ac}{4a^2}\ge0

. If

b^2-4ac<0

, a square would have to equal a negative number, which is impossible for real

x

: no real roots. The graph never meets the

x

-axis.

Completing the square: the procedure

Drawing diagram…

The same route works for every quadratic. The only branch is whether you first need to take out $a$ .

7. Using the completed square: back to the ball

Worked exampleMaximum height of the ball, $h=1+20t-5t^2$

A ball's height is $h=1+20t-5t^2$ metres after $t$ seconds. Find the greatest height and the time it happens.

Goal: Write $h$ as $a(t+p)^2+q$ ; then $q$ is the maximum (since $a<0$ ) and $t=-p$ is when.
Strategy: Reorder in powers of $t$ , take out $a=-5$ , complete the square inside.

$h=-5t^2+20t+1$
Reorder: $a=-5$ , $b=20$ , $c=1$ .
$=-5\left(t^2-4t\right)+1$
Factor $-5$ : $20t=-5\times(-4t)$ . Careful with the sign.
$=-5\left[(t-2)^2-4\right]+1$
Halve $-4$ : $-2$ ; square: $4$ ; take away.
$=-5(t-2)^2+20+1$
$-5\times(-4)=+20$ .
$=-5(t-2)^2+21$
Simplify.
$-5(t-2)^2\le0\;\Rightarrow\;h\le21$
A square is $\ge0$ ; times $-5$ it's $\le0$ .

AnswerMaximum height

21

m, at

t=2

s exactly.

✓ Check

h(2)=1+40-20=21

✓. Symmetry:

h(1)=h(3)=16

, equal values either side of

t=2

✓.

Worked exampleSolving by completing the square: $x^2-4x-1=0$

Solve $x^2-4x-1=0$ , giving exact answers.

Goal: Find all real $x$ , in surd form.
Strategy: Complete the square so $x$ appears once, isolate the square, square-root both sides.

$(x-2)^2-4-1=0$
Halve $-4$ : $-2$ ; square: $4$ ; take away.
$(x-2)^2=5$
Add $5$ to both sides.
$x-2=\pm\sqrt5$
Square root both sides, both signs.
$x=2\pm\sqrt5$
Add $2$ .

Answer

x=2+\sqrt5\approx4.24

or

x=2-\sqrt5\approx-0.24

.

✓ Check

x=2+\sqrt5

:

x^2=9+4\sqrt5

,

4x=8+4\sqrt5

, so

x^2-4x-1=9+4\sqrt5-8-4\sqrt5-1=0

✓.

Key takeawayThree things the completed square gives you (1) the vertex and line of symmetry; (2) the least or greatest value (and so the range of a function); (3) the solutions via a single square root.

Pause and think

What is the range of $f(x)=x^2+8x+20$ ?

Summary

$(x+p)^2=x^2+2px+p^2$ , proved by expanding.
$x^2+bx=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2$ : halve, square, take away.
For $a\neq1$ , factor $a$ out of the $x$ -terms first, then multiply it back into everything.
$a(x+p)^2+q$ : vertex $(-p,q)$ ; min if $a>0$ , max if $a<0$ .
The quadratic formula is completing the square applied to $ax^2+bx+c=0$ .

Completing the square: Rewriting $ax^2+bx+c$ as $a(x+p)^2+q$ .
Vertex: The turning point of a parabola.
Line of symmetry: The vertical mirror line through the vertex.
Perfect square: An expression of the form $(x+p)^2$ .
Coefficient: The number multiplying a power of the variable.

HistoryAlgebra was named after this method. Around 820 CE, the Baghdad scholar Muhammad ibn Musa al-Khwarizmi wrote a book whose title contains the word al-jabr ("restoring"), which gave us the word algebra; his own name gave us algorithm. One of his worked problems is

x^2+10x=39

, which he solved by drawing a square and literally completing it with a

5\times5

corner, getting

x=3

. You will do the same calculation in symbols.

Fun factOlder than algebra itself. Babylonian clay tablets from about 1800 BCE already contain quadratic problems about areas of fields, solved by a recipe equivalent to completing the square. Quadratics are nearly 4000 years old.

Section 3 of 14

Key formulae

a = 1: halve, square, subtract

x^2+bx+c=\left(x+\frac b2\right)^2-\left(\frac b2\right)^2+c

Any a: take a out of the x terms first

ax^2+bx+c=a\left[x^2+\frac bax\right]+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}

Reading the graph from y = a(x + p)² + q

\begin{gathered}\text{vertex }(-p,\ q)\qquad\text{line of symmetry }x=-p \\[3pt] a>0:\ \text{minimum value }q\qquad a<0:\ \text{maximum value }q\end{gathered}

NoteThe sign flips.

(x-3)^2

is zero when

x=+3

, so

y=(x-3)^2+5

has its vertex at

(3,\,5)

. The

x

-coordinate of the vertex is always the opposite sign of the number in the bracket.

Section 4 of 14

Worked examples

Example 1 · a = 10 / 3 steps

Express

x^2+8x+3

in the form

(x+p)^2+q

.

Example 2 · a ≠ 1 (exam style)0 / 4 steps

f(x)=2x^2-12x+23

. (a) Express

f(x)

in the form

a(x+b)^2+c

. (b) State the vertex of

y=f(x)

. (c) State the range of

f

.

Example 3 · Negative a gives a maximum0 / 3 steps

Express

5+4x-x^2

in the form

a-(x-b)^2

and state its greatest value.

Example 4 · Working backwards from the vertex0 / 2 steps

The curve

y=x^2+px+q

has its minimum point at

(3,\,-7)

. Find

p

and

q

.

Example 5 · In context0 / 2 steps

A ball is thrown upwards. Its height after

t

seconds is

h=20t-5t^2

metres. Find its greatest height and when it is reached.

Section 5 of 14

Explore the graph

The vertex form in action

Loading the interactive graph…

Try this

Move $h$ . The vertex slides left and right, and the line of symmetry moves with it. Notice $h=2$ puts it at $x=+2$ : that is $(x-2)^2$ .
Make $a$ negative. What happens to the shape, and is $(h,k)$ now a minimum or a maximum?
Set $a=1$ and move $k$ up from $-3$ . At what value of $k$ do the two roots join into one? What happens after that?

$y=a(x-h)^2+k$ always has its vertex at $(h,\,k)$ . The roots disappear exactly when the vertex lifts off the $x$ -axis, which is the discriminant changing sign (next subtopic but one).

Prefer dragging? In this explorer you can grab the vertex itself and watch the equation update.

a 1h 2k -3

Completed square: $y=(x-2)^2-3$
Expanded: $y=x^2-4x+1$
Vertex: $(2,\ -3)$ minimum
Line of symmetry: $x=2$
Roots: $x=0.268,\ 3.732$
Range: $y\ge -3$

TrickHalve, square, subtract. Say it as you write: "half of

8

is

4

, bracket

(x+4)

squared, subtract

4^2=16

". For

a\ne1

, add one step at the start: "take

a

out of the

x

terms", and one at the end: "multiply the subtracted square by

a

".

In the real worldWhy parabolas are everywhere. Ignoring air resistance, anything thrown follows a parabola, so its greatest height is the vertex. Satellite dishes and car headlights are parabola-shaped too: every ray parallel to the axis bounces off the curve to a single point, the focus.

Section 6 of 14

Common mistakes

Incorrect

x^2+6x+1=(x+3)^2+9+1

Correct

(x+3)^2-9+1

(x+3)^2

already contains

+9

. Subtract it to keep the expression the same.

Incorrect

2x^2-12x+23=2(x-3)^2-9+23

Correct

2\left[(x-3)^2-9\right]+23=2(x-3)^2+5

The

-9

is inside the bracket multiplied by

2

.

Incorrect

y=(x-3)^2+5

has vertex

(-3,\,5)

Correct

(3,\,5)

(x-3)^2=0

when

x=+3

.

Incorrect

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

Correct

9-(x-2)^2

-(x^2-4x)=-\left[(x-2)^2-4\right]=-(x-2)^2+4

: the minus sign turns

-4

into

+4

.

Expectation

$(x+3)^2=x^2+9$

Reality

$(x+3)^2=x^2+6x+9$ , and the $6x$ has been in the room the whole time

Section 7 of 14

Exam tip

Exam tip"Hence" means read it off. If part (a) asks for

a(x+b)^2+c

and part (b) says "hence state the least value" or "hence state the coordinates of the vertex", the answer is already on your page:

c

, and

(-b,\,c)

. One line, one mark. Do not differentiate or start again.

Section 8 of 14

Guided practice

SA-9709-P1-QD-G013 marksL1 FoundationCalculatora = 1

x^2-4x+9=(x+p)^2+q

. Work through it in three steps.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P1-QD-0012 marksL1 FoundationCalculatora = 1

Express

x^2+6x+11

in the form

(x+a)^2+b

, where

a

and

b

are constants.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD-0033 marksL2 StandardCalculatora = 1

Express

x^2+5x-3

in the form

(x+a)^2+b

, giving

a

and

b

as exact fractions.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD-0043 marksL2 StandardCalculatora ≠ 1

Express

2x^2-12x+5

in the form

a(x+b)^2+c

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD-0052 marksL2 StandardCalculatorVertex

Find the coordinates of the minimum point of the curve

y=x^2-8x+19

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD-0114 marksL4 ChallengingCalculatora ≠ 1

Express

5+4x-x^2

in the form

a-(x-b)^2

, and hence state the maximum value of

5+4x-x^2

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD-0124 marksL4 ChallengingCalculatorUnknown constants

The curve

y=x^2+px+q

has a minimum point at

(3,\,-7)

. Find

p

and

q

.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD1-0014 marksL2 StandardCalculatorSketching & models

For the curve

y=x^2-2x-8

, find (a) the

x

-coordinates where it crosses the

x

-axis (b) where it crosses the

y

-axis (c) the coordinates of its vertex. (These are what a sketch must show.)

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-QD1-0054 marksL3 ExamCalculatorSketching & models

A farmer has

40

m of fencing to make three sides of a rectangular pen against a long straight wall. The two sides perpendicular to the wall are each

x

m long. (a) Show that the area is

A=x(40-2x)

. (b) By completing the square, find the greatest possible area.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P1-QD-0135 marksL5 ExtensionCalculatorUnknown constants

The curve

y=x^2-2kx+3k

, where

k

is a positive constant, has its minimum point on the

x

-axis. Find

k

.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Halve the coefficient of $x$ , square the bracket, subtract the square. For $a\ne1$ , take $a$ out of the $x$ terms first.
$y=a(x+p)^2+q$ has vertex $(-p,\,q)$ and line of symmetry $x=-p$ .
$a>0$ : least value $q$ . $a<0$ : greatest value $q$ .
To sketch: vertex, $y$ -intercept ( $x=0$ ), and roots if there are any.
In context, the vertex answers "greatest height", "maximum area" or "least cost".

Section 12 of 14

Quick quiz

Q1 · SA-9709-P1-QD-Q011 markL1 FoundationCalculatora = 1

x^2+10x

is equal to

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P1-QD-Q021 markL1 FoundationCalculatorVertex

The vertex of

y=(x+2)^2-7

is

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P1-QD-Q031 markL2 StandardCalculatorUnknown constants

Which form shows directly that

x^2-6x+11>0

for all

x

?

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 7357 A-level Mathematics
Where: Section B, Algebra and functions (B3–B5)
Expected: Completing the square, including to find the turning point of a quadratic and to sketch its graph.
Calculator: Permitted on all papers.
Typical style: AQA sometimes asks you to explain a step in words, or to set a quadratic in a real-world context.
Terminology: AQA and Edexcel call a repeated root "equal roots" as often as "a repeated root".
Open in: 7357 · Paper 1 7357 · Paper 2 7357 · Paper 3

Section 14 of 14

Past paper questions

6 questions (42 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/12 · Oct/Nov 2025Question 15 marks
Complete the square, the condition for no real roots, and an exact solution.
Question paper, page 2 Mark scheme
9709/12 · May/June 2025Question 119 marks
Complete the square, then find an inverse and a composite function.
Question paper, page 16 Mark scheme
9709/12 · Oct/Nov 2024Question 78 marks
Complete the square to find a vertex, then the area between a parabola and a line.
Question paper, page 10 Mark scheme
9709/12 · Oct/Nov 2023Question 68 marks
Minimum point of a quadratic, then its image after a stretch and a translation.
Question paper, page 8 Mark scheme
9709/12 · May/June 2023Question 33 marks
Complete the square, then use the discriminant for a quadratic with no real roots.
Question paper, page 4 Mark scheme
9709/12 · Oct/Nov 2022Question 69 marks
Complete the square, solve with it, and sketch the turning point.
Question paper, page 8 Mark scheme

All 9709 past paper questions by topic