The temperature in a greenhouse, T degrees Celsius, t hours after midnight is modelled by T = t^2 - 12t + 40 for 0 t 12. Find the time at which the temperature is lowest, and that lowest temperature.

t (hours) = 6, lowest temperature (degrees C) = 4

How many marks is completing the square worth on Edexcel Higher?

The form itself is usually 2 marks when the coefficient of x^2 is 1 and 3 marks when it is not. A follow-on part asking for the turning point, the line of symmetry or the minimum value adds 1-2 marks, so the whole question is commonly 3-5 marks.

Do I have to complete the square if the question just says solve?

Only if the stem says so. If it says 'by completing the square' or 'hence solve', the mark scheme awards the method marks for that route alone, so the quadratic formula will not score. If the wording is plain 'solve', any correct method is accepted.

Why does the turning point have the opposite sign to the number in the bracket?

A square is never negative, so (x + p)^2 + q takes its smallest value when the bracket is exactly zero, and that happens at x = -p. Substituting gives y = q, so the turning point is (-p, q) and the line of symmetry is x = -p.

Completing the Square - Edexcel GCSE Higher Maths

Typically 3-5 marks when it appears, often split across two parts · Spec 1MA1

Completing the square rewrites a quadratic as a squared bracket plus a number: $x^2 + bx + c$ becomes $(x + p)^2 + q$ , and $ax^2 + bx + c$ becomes $a(x + p)^2 + q$ . On Edexcel's 1MA1 it is Higher tier only. It sits under spec ref A18, solving quadratic equations by completing the square, and A11, deducing the turning point and the roots of a quadratic from its completed square form. Foundation candidates solve quadratics by factorising and are never asked for this.

Edexcel nearly always splits the question into parts. Part (a) asks for the form and carries 2-3 marks. Part (b) opens with 'Hence write down' and wants the coordinates of the turning point, the line of symmetry or the minimum value, for 1-2 marks. The word hence is an instruction, not a hint: a turning point produced by any other route scores nothing in part (b). Papers 2 and 3 allow a calculator, but when the stem says 'give your solutions in surd form' or 'find the exact solutions', the calculator cannot supply the answer anyway, so practise the algebra by hand.

Three slips account for most of the lost marks. The compensating $-p^2$ gets dropped, so $x^2 + 6x + 1$ comes back as $(x + 3)^2 + 1$ instead of $(x + 3)^2 - 8$ . When the coefficient of $x^2$ is not 1, the number pulled outside the bracket is never multiplied back through. And the turning point is written with the sign of $p$ rather than $-p$ , turning $(-3, -13)$ into $(3, -13)$ . Expanding your own bracket back out takes ten seconds and catches all three before the examiner does.

Worked example

(a) Write

2x^2 + 12x + 5

in the form

a(x + p)^2 + q

, where

a

p

and

q

are integers. (b) Hence write down the coordinates of the turning point of the curve

y = 2x^2 + 12x + 5

[4 marks]

Take the factor of 2 out of the $x^2$ and $x$ terms only, leaving the 5 outside: $2[x^2 + 6x] + 5$ .
M1: factorising 2 from the first two terms only, not from the constant
Complete the square inside the bracket. Half of 6 is 3, so $x^2 + 6x = (x + 3)^2 - 9$ , giving $2[(x + 3)^2 - 9] + 5$ .
M1: correct $(x + 3)^2 - 9$ inside the bracket
Multiply the whole bracket by 2 and collect the numbers: $2(x + 3)^2 - 18 + 5 = 2(x + 3)^2 - 13$ .
A1: $a = 2$ , $p = 3$ , $q = -13$ ; the $-9$ must be doubled to $-18$
$2(x + 3)^2$ is smallest when the bracket is zero, which happens at $x = -3$ , and then $y = -13$ . Turning point $(-3, -13)$ .
B1 (follow through from part a): read the $x$ -coordinate as $-p$ , not $p$

Practice questions

These are original questions written in Edexcel 1MA1 Higher style. They run through the same sequence Edexcel uses: put the quadratic into completed square form, then read off the turning point, the line of symmetry or the minimum value, and finally solve with surds. Fractional values of $p$ appear, as they do on the real papers, and you can type answers like 7/2 or 3.5.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Exam pace [2 marks]

Write $x^2 - 3x + 5$ in the form $(x + p)^2 + q$ . Find the value of $p$ and the value of $q$ .

p = q =
Question 2 Exam pace [2 marks]

Write $x^2 + x - 1$ in the form $(x + p)^2 + q$ . Find the value of $p$ and the value of $q$ .

p = q =
Question 3 Exam pace [3 marks]

Writing $x^2 - 8x + 20$ in the form $(x + p)^2 + q$ proves that the expression is positive for every value of $x$ . Find the value of $p$ and the value of $q$ .

p = q =
Question 4 Exam pace [4 marks]

The temperature in a greenhouse, $T$ degrees Celsius, $t$ hours after midnight is modelled by $T = t^2 - 12t + 40$ for $0 \le t \le 12$ . Find the time at which the temperature is lowest, and that lowest temperature.

t (hours) = lowest temperature (degrees C) =
Question 5 Stretch [1 mark]

Find the exact solutions of $x^2 - 3x - 1 = 0$ by completing the square. Which pair of solutions is correct?

Answer options for question 5
$x = \frac{3 \pm \sqrt{13}}{2}$ $x = \frac{-3 \pm \sqrt{13}}{2}$ $x = \frac{3 \pm \sqrt{5}}{2}$ $x = 3 \pm \frac{\sqrt{13}}{2}$
Question 6 Stretch [1 mark]

Find the exact solutions of $x^2 + 5x + 2 = 0$ by completing the square. Which pair of solutions is correct?

Answer options for question 6
$x = \frac{-5 \pm \sqrt{17}}{2}$ $x = \frac{5 \pm \sqrt{17}}{2}$ $x = \frac{-5 \pm \sqrt{33}}{2}$ $x = -5 \pm \frac{\sqrt{17}}{2}$

Common mistakes examiners see

Writing the bracket and stopping, so $x^2 + 10x + 3$ is given as $(x + 5)^2 + 3$ . Expanding that gives $x^2 + 10x + 28$ , which is not the expression you were handed.

After halving the $x$ coefficient to get $p$ , always subtract $p^2$ : $x^2 + 10x + 3 = (x + 5)^2 - 25 + 3 = (x + 5)^2 - 22$ . Expand your final answer to check it returns the original.
With $a \ne 1$ , forgetting to multiply the compensating term back through, so $2x^2 + 12x + 5$ becomes $2(x + 3)^2 - 9 + 5 = 2(x + 3)^2 - 4$ instead of $2(x + 3)^2 - 13$ .

Write the middle line with square brackets, $2[(x + 3)^2 - 9] + 5$ , and treat the multiplication as a separate line of working. Everything inside the bracket gets doubled, including the $-9$ .
Reading the turning point straight off the signs, so $(x + 4)^2 - 7$ is reported as $(4, -7)$ .

The bracket is zero when $x + 4 = 0$ , so $x = -4$ . Say 'opposite sign for $x$ , same sign for $y$ ' as you write the coordinates: $(-4, -7)$ .
Halving after squaring, or halving the constant term instead of the $x$ coefficient, so $x^2 + 7x + 1$ turns into $(x + 3)^2$ or $(x + 7)^2$ working.

Halve the number attached to $x$ first, and keep it as a fraction: for $x^2 + 7x + 1$ , $p = \frac{7}{2}$ and the form is $(x + \frac{7}{2})^2 - \frac{45}{4}$ . Edexcel sets odd coefficients precisely to see whether you can handle the fraction.
Solving $(x + 3)^2 = 5$ by writing $x + 3 = \sqrt{5}$ and giving one answer, or rounding it on a calculator paper when the question asked for exact solutions.

Square rooting gives $\pm$ , so $x = -3 \pm \sqrt{5}$ : two solutions, left in surd form. If the stem says 'exact', a decimal scores nothing however accurate it is.

Frequently asked questions

Is completing the square on the Foundation paper?: No. In Edexcel 1MA1 it is Higher tier only, under spec refs A18 and A11. Foundation candidates solve quadratics by factorising and by reading a graph, and are not asked to produce the $(x + p)^2 + q$ form.
How many marks is completing the square worth on Edexcel Higher?: The form itself is usually 2 marks when the coefficient of $x^2$ is 1 and 3 marks when it is not. A follow-on part asking for the turning point, the line of symmetry or the minimum value adds 1-2 marks, so the whole question is commonly 3-5 marks.
Do I have to complete the square if the question just says solve?: Only if the stem says so. If it says 'by completing the square' or 'hence solve', the mark scheme awards the method marks for that route alone, so the quadratic formula will not score. If the wording is plain 'solve', any correct method is accepted.
Why does the turning point have the opposite sign to the number in the bracket?: A square is never negative, so $(x + p)^2 + q$ takes its smallest value when the bracket is exactly zero, and that happens at $x = -p$ . Substituting gives $y = q$ , so the turning point is $(-p, q)$ and the line of symmetry is $x = -p$ .

Completing the Square - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics