Mathematics 9709 · Pure Mathematics 1

Completing the square

Completing the square rewrites $ax^2+bx+c$ as $a(x-h)^2+k$ . Now $x$ appears only once.
From this one change you can read the vertex, the least value, and the range. You do not need calculus.

Why it works

Build it, do not memorise it. Squaring the bracket gives back $x^2+bx$ plus one extra bit, $(\tfrac b2)^2$ . So you take that bit away again.

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

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

Halve $b$ for the bracket, then subtract its square.
Check with $b=6$ : $(x+3)^2=x^2+6x+9$ , so $x^2+6x=(x+3)^2-9$ .

The method: halve, square, subtract

For $x^2+bx+c$ : halve $b$ for the bracket. Subtract $(\tfrac b2)^2$ . Then bring down the $c$ and tidy up.

Worked example

Write $x^2+6x+5$ in completed-square form

Halve $6$ → bracket $(x+3)^2$ .
Subtract $3^2=9$ , carry the $+5$ : $(x+3)^2-9+5$ .
Tidy $-9+5$ .

Answer

(x+3)^2-4

Common mistake

The sign in the bracket follows

b

. But the number you subtract is always positive. For

x^2-10x+7

: half of

-10

-5

, and

(-5)^2=+25

, since squaring turns the minus into a plus. So you get

(x-5)^2-25+7=(x-5)^2-18

Your turn— tap to reveal the worked answer (quick check)

Complete the square on $x^2+10x+9$ : half of $10$ is $5$ , subtract $25$ , carry $+9$ .

Answer

(x+5)^2-16

When a ≠ 1

If $a\neq 1$ , take $a$ out of the first two terms only. Complete the square inside the bracket. Then multiply $a$ back in.

Worked example

Write $2x^2-12x+3$ in the form $a(x-h)^2+k$

Factor $2$ from the first two terms: $2(x^2-6x)+3$ .
Inside: half of $-6$ is $-3$ , subtract $9$ : $2\big[(x-3)^2-9\big]+3$ .
Multiply the $2$ through — the $-9$ becomes $-18$ : $2(x-3)^2-18+3$ .
Tidy.

Answer

2(x-3)^2-15

Common mistake

When you multiply the factor back in, it must reach both terms in the bracket. In

2\big[(x-3)^2-9\big]

the

-9

becomes

-18

, not

-9

. This is the most common mistake in the whole topic.

When the form keeps the factor OUTSIDE (9709/12 Mar 2022 Q5)

“Express $2x^2-8x+14$ in the form $2\big[(x-a)^2+b\big]$ .” Here the $2$ must stay outside. So do not multiply through. Instead, move the constant into the bracket.

Factor $2$ from the first two terms, carry $+14$ : $2(x^2-4x)+14$ .
Inside: half of $-4$ is $-2$ , subtract $4$ : $2\big[(x-2)^2-4\big]+14$ .
Pull $+14$ inside as $+7$ (since $14=2\times7$ ): $2\big[(x-2)^2-4+7\big]$ .

Answer

2\big[(x-2)^2+3\big]

Reading the vertex

From $a(x-h)^2+k$ , you can read the turning point at once: $(h,\,k)$ . You do not need calculus.

A square is never negative. So (for $a>0$ ) the smallest value is $k$ . You reach it when the bracket is $0$ , that is, at $x=h$ .
So the turning point is $(h,k)$ . The curve is symmetric about $x=h$ . From $2(x-3)^2-15$ the minimum is $(3,-15)$ .

y = (x − 1)² + 2·vertex (1, 2)

Drag h and k, and the vertex moves to match the completed-square form.

Common mistake

The

x

-value of the vertex changes sign.

(x-3)^2

turns at

x=+3

, but

(x+3)^2

turns at

x=-3

. So

(x+5)^2-16

has its minimum at

(-5,\,-16)

Your turn— tap to reveal the worked answer (9709/11 Jun 2022 Q1)

Express $x^2-6x+2$ in completed-square form and state its least value. Half of $-6$ is $-3$ , subtract $9$ , carry $+2$ : $(x-3)^2-9+2$ .

Answer

(x-3)^2-7;\ \text{ least value } -7 \text{ at } x=3

Range and least value

The least value $k$ is the bottom of the range. For an upward parabola the range is $f(x)\ge k$ . Write it in $f(x)$ , never in $x$ .

Worked example

$-2x^2-8x-13$ : complete the square, then give the range

Opens downward; factor out $-2$ , watching signs ( $-8x\div-2=+4x$ ): $-2(x^2+4x)-13$ .
Inside: half of $4$ is $2$ , subtract $4$ : $-2\big[(x+2)^2-4\big]-13$ .
Multiply $-2$ through ( $-2\times-4=+8$ ) and tidy: $-2(x+2)^2-5$ . Maximum value $-5$ .

Answer

f(x)\le -5

Common mistake

Sometimes the question limits the domain. Then the range may not reach the vertex. Check if the

x

of the vertex is inside the domain. If it is not, use the end value of the domain instead. (9709/13 Nov 2022 Q2)

Limited domain: the vertex is outside it (9709/13 Nov 2022 Q2)

Take $-2(x+2)^2-5$ on the domain $x<-3$ . The vertex is at $x=-2$ , which is outside the domain. So the curve never reaches its highest point. Test the end value $x=-3$ : $f(-3)=-2(1)-5=-7$ . For $x<-3$ the downward curve only goes lower, so:

Answer

f(x)<-7

When to use it

If it says “Express in the form $a(x\pm h)^2\pm k$ ”, or asks for the vertex / turning point / minimum / maximum / least value / range of a quadratic, then complete the square.

If the question wants…	Reach for instead
an exact / surd solution of a quadratic	the formula (§1.3) — cleaner for surds
to prove it is always positive / has no real roots	the discriminant (§1.5)

Checking on a calculator

“Express in the form” gives marks for shown algebra. An answer with no working scores nothing. Here the calculator only checks your work. It does not replace the method.

Substitute $x=h$ into the original quadratic — you should get exactly $k$ .
Or expand your answer back on the calculator — it must return the original $ax^2+bx+c$ .

PreviousSolving by factorising NextThe quadratic formula

Completing the square

Completing the square rewrites $ax^2+bx+c$ as $a(x-h)^2+k$ . Now $x$ appears only once.
From this one change you can read the vertex, the least value, and the range. You do not need calculus.

Why it works

Build it, do not memorise it. Squaring the bracket gives back $x^2+bx$ plus one extra bit, $(\tfrac b2)^2$ . So you take that bit away again.

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

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

Halve $b$ for the bracket, then subtract its square.
Check with $b=6$ : $(x+3)^2=x^2+6x+9$ , so $x^2+6x=(x+3)^2-9$ .

The method: halve, square, subtract

For $x^2+bx+c$ : halve $b$ for the bracket. Subtract $(\tfrac b2)^2$ . Then bring down the $c$ and tidy up.

Worked example

Write $x^2+6x+5$ in completed-square form

Halve $6$ → bracket $(x+3)^2$ .
Subtract $3^2=9$ , carry the $+5$ : $(x+3)^2-9+5$ .
Tidy $-9+5$ .

Answer

(x+3)^2-4

Common mistake

The sign in the bracket follows

b

. But the number you subtract is always positive. For

x^2-10x+7

: half of

-10

-5

, and

(-5)^2=+25

, since squaring turns the minus into a plus. So you get

(x-5)^2-25+7=(x-5)^2-18

Your turn— tap to reveal the worked answer (quick check)

Complete the square on $x^2+10x+9$ : half of $10$ is $5$ , subtract $25$ , carry $+9$ .

Answer

(x+5)^2-16

When a ≠ 1

If $a\neq 1$ , take $a$ out of the first two terms only. Complete the square inside the bracket. Then multiply $a$ back in.

Worked example

Write $2x^2-12x+3$ in the form $a(x-h)^2+k$

Factor $2$ from the first two terms: $2(x^2-6x)+3$ .
Inside: half of $-6$ is $-3$ , subtract $9$ : $2\big[(x-3)^2-9\big]+3$ .
Multiply the $2$ through — the $-9$ becomes $-18$ : $2(x-3)^2-18+3$ .
Tidy.

Answer

2(x-3)^2-15

Common mistake

When you multiply the factor back in, it must reach both terms in the bracket. In

2\big[(x-3)^2-9\big]

the

-9

becomes

-18

, not

-9

. This is the most common mistake in the whole topic.

When the form keeps the factor OUTSIDE (9709/12 Mar 2022 Q5)

“Express $2x^2-8x+14$ in the form $2\big[(x-a)^2+b\big]$ .” Here the $2$ must stay outside. So do not multiply through. Instead, move the constant into the bracket.

Factor $2$ from the first two terms, carry $+14$ : $2(x^2-4x)+14$ .
Inside: half of $-4$ is $-2$ , subtract $4$ : $2\big[(x-2)^2-4\big]+14$ .
Pull $+14$ inside as $+7$ (since $14=2\times7$ ): $2\big[(x-2)^2-4+7\big]$ .

Answer

2\big[(x-2)^2+3\big]

Reading the vertex

From $a(x-h)^2+k$ , you can read the turning point at once: $(h,\,k)$ . You do not need calculus.

A square is never negative. So (for $a>0$ ) the smallest value is $k$ . You reach it when the bracket is $0$ , that is, at $x=h$ .
So the turning point is $(h,k)$ . The curve is symmetric about $x=h$ . From $2(x-3)^2-15$ the minimum is $(3,-15)$ .

y = (x − 1)² + 2·vertex (1, 2)

Drag h and k, and the vertex moves to match the completed-square form.

Common mistake

The

x

-value of the vertex changes sign.

(x-3)^2

turns at

x=+3

, but

(x+3)^2

turns at

x=-3

. So

(x+5)^2-16

has its minimum at

(-5,\,-16)

Your turn— tap to reveal the worked answer (9709/11 Jun 2022 Q1)

Express $x^2-6x+2$ in completed-square form and state its least value. Half of $-6$ is $-3$ , subtract $9$ , carry $+2$ : $(x-3)^2-9+2$ .

Answer

(x-3)^2-7;\ \text{ least value } -7 \text{ at } x=3

Range and least value

The least value $k$ is the bottom of the range. For an upward parabola the range is $f(x)\ge k$ . Write it in $f(x)$ , never in $x$ .

Worked example

$-2x^2-8x-13$ : complete the square, then give the range

Opens downward; factor out $-2$ , watching signs ( $-8x\div-2=+4x$ ): $-2(x^2+4x)-13$ .
Inside: half of $4$ is $2$ , subtract $4$ : $-2\big[(x+2)^2-4\big]-13$ .
Multiply $-2$ through ( $-2\times-4=+8$ ) and tidy: $-2(x+2)^2-5$ . Maximum value $-5$ .

Answer

f(x)\le -5

Common mistake

Sometimes the question limits the domain. Then the range may not reach the vertex. Check if the

x

of the vertex is inside the domain. If it is not, use the end value of the domain instead. (9709/13 Nov 2022 Q2)

Limited domain: the vertex is outside it (9709/13 Nov 2022 Q2)

Answer

f(x)<-7

When to use it

If it says “Express in the form $a(x\pm h)^2\pm k$ ”, or asks for the vertex / turning point / minimum / maximum / least value / range of a quadratic, then complete the square.

If the question wants…	Reach for instead
an exact / surd solution of a quadratic	the formula (§1.3) — cleaner for surds
to prove it is always positive / has no real roots	the discriminant (§1.5)

Checking on a calculator

“Express in the form” gives marks for shown algebra. An answer with no working scores nothing. Here the calculator only checks your work. It does not replace the method.

Substitute $x=h$ into the original quadratic — you should get exactly $k$ .
Or expand your answer back on the calculator — it must return the original $ax^2+bx+c$ .