Using the Completed Square Form | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

The forms the exam asks for

Match the form named in the question. The square is the same; only the final arrangement changes. Letters in the requested form stand for constants you must find; match their positions, not their names.

Form	Last-step cue
$(x+a)^2+b$	plain completed-square form
$a(x+b)^2+c$	take the x² coefficient out first
$a-(x+b)^2$	take out a negative factor
$(ax+b)^2+c$	match the square directly
$2\big[(x-a)^2+b\big]$	move the final constant inside

Form: $(x+a)^2+b$
Last-step cue: plain completed-square form

Form: $a(x+b)^2+c$
Last-step cue: take the x² coefficient out first

Form: $a-(x+b)^2$
Last-step cue: take out a negative factor

Form: $(ax+b)^2+c$
Last-step cue: match the square directly

Form: $2\big[(x-a)^2+b\big]$
Last-step cue: move the final constant inside

Worked example

Exam variant: the (ax + b)² form

Express $4x^2-12x+13$ as $(2x+a)^2+b$ (9709/11/O/N/23 Q9(a)).

(2x+a)^2=4x^2+4ax+a^2

4a=-12 \;\Rightarrow\; a=-3

Answer

(2x-3)^2+4

Worked example

Exam variant: keep the factor outside the bracket

Express $2x^2-8x+14$ as $2\big[(x-a)^2+b\big]$ (9709/12/F/M/22 Q5(a)).

2\big[(x-2)^2-4\big]+14

=2\big[(x-2)^2-4+7\big]

Answer

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

Your turn9709/12/M/J/23 Q3(a)

Express $4x^2-24x+p$ in the form $a(x+b)^2+c$ , where $a$ and $b$ are integers and $c$ is given in terms of $p$ .

Show worked answer

The letter $p$ changes nothing. Carry it along like any other constant.

4(x^2-6x)+p

=4\big[(x-3)^2-9\big]+p

=4(x-3)^2-36+p

Answer

4(x-3)^2+(p-36)

Solving an equation this way

Reuse the completed square. Undo a square with $\pm\sqrt{\ }$ and keep the answer exact when asked.

Worked example

One quadratic, three questions (9709/12/O/N/25 Q1)

Complete the square once, then reuse the result in every part.

(a) Express $9x^2-36x+8$ in the form $p(x+q)^2+r$ . Take $9$ out of the first two terms, complete the square inside, then multiply back in.
$9(x^2-4x)+8$
$=9\big[(x-2)^2-4\big]+8$
$=9(x-2)^2-28$
(b) Find the values of $k$ for which $9x^2-36x+8=k$ has no real roots. Use part (a) and rearrange.
$9(x-2)^2-28=k$
$9(x-2)^2=k+28$
The left side is a square multiplied by $9$ , so it can never be negative. There is no solution when the right side is negative.
$k+28<0 \;\Rightarrow\; k<-28$
(c) Find the exact roots of $9x^2-36x+8=-15$ . Use part (a) again, then undo the square.
$9(x-2)^2=13$
$(x-2)^2=\tfrac{13}{9}$
$x-2=\pm\tfrac{\sqrt{13}}{3}$

Answer

\text{(a) } 9(x-2)^2-28

\text{(b) } k<-28

\text{(c) } x=\tfrac{6\pm\sqrt{13}}{3}

The forms the exam asks for

Form	Last-step cue
$(x+a)^2+b$	plain completed-square form
$a(x+b)^2+c$	take the x² coefficient out first
$a-(x+b)^2$	take out a negative factor
$(ax+b)^2+c$	match the square directly
$2\big[(x-a)^2+b\big]$	move the final constant inside

Form: $(x+a)^2+b$
Last-step cue: plain completed-square form

Form: $a(x+b)^2+c$
Last-step cue: take the x² coefficient out first

Form: $a-(x+b)^2$
Last-step cue: take out a negative factor

Form: $(ax+b)^2+c$
Last-step cue: match the square directly

Form: $2\big[(x-a)^2+b\big]$
Last-step cue: move the final constant inside

Worked example

Exam variant: the (ax + b)² form

Express $4x^2-12x+13$ as $(2x+a)^2+b$ (9709/11/O/N/23 Q9(a)).

(2x+a)^2=4x^2+4ax+a^2

4a=-12 \;\Rightarrow\; a=-3

Answer

(2x-3)^2+4

Worked example

Exam variant: keep the factor outside the bracket

Express $2x^2-8x+14$ as $2\big[(x-a)^2+b\big]$ (9709/12/F/M/22 Q5(a)).

2\big[(x-2)^2-4\big]+14

=2\big[(x-2)^2-4+7\big]

Answer

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

Your turn9709/12/M/J/23 Q3(a)

Express $4x^2-24x+p$ in the form $a(x+b)^2+c$ , where $a$ and $b$ are integers and $c$ is given in terms of $p$ .

Show worked answer

The letter $p$ changes nothing. Carry it along like any other constant.

4(x^2-6x)+p

=4\big[(x-3)^2-9\big]+p

=4(x-3)^2-36+p

Answer

4(x-3)^2+(p-36)

Solving an equation this way

Reuse the completed square. Undo a square with $\pm\sqrt{\ }$ and keep the answer exact when asked.

Worked example

One quadratic, three questions (9709/12/O/N/25 Q1)

Complete the square once, then reuse the result in every part.

(a) Express $9x^2-36x+8$ in the form $p(x+q)^2+r$ . Take $9$ out of the first two terms, complete the square inside, then multiply back in.
$9(x^2-4x)+8$
$=9\big[(x-2)^2-4\big]+8$
$=9(x-2)^2-28$
(b) Find the values of $k$ for which $9x^2-36x+8=k$ has no real roots. Use part (a) and rearrange.
$9(x-2)^2-28=k$
$9(x-2)^2=k+28$
The left side is a square multiplied by $9$ , so it can never be negative. There is no solution when the right side is negative.
$k+28<0 \;\Rightarrow\; k<-28$
(c) Find the exact roots of $9x^2-36x+8=-15$ . Use part (a) again, then undo the square.
$9(x-2)^2=13$
$(x-2)^2=\tfrac{13}{9}$
$x-2=\pm\tfrac{\sqrt{13}}{3}$

Answer

\text{(a) } 9(x-2)^2-28

\text{(b) } k<-28

\text{(c) } x=\tfrac{6\pm\sqrt{13}}{3}