- Completing the square rewrites a quadratic \( ax^2 + bx + c \) as \( a(x-h)^2 + k \), called the vertex form.
- Because a square is never negative, the vertex form shows the maximum or minimum at a glance: the turning point is \( (h, k) \).
- Case: a theatre selling 400 tickets at $20, losing 10 sales per $1 rise, has revenue \( R = -10(x-10)^2 + 9000 \). The best price is $30, for $9,000.
- The same move solves quadratics, gives the quadratic formula, and turns a messy circle equation into centre-and-radius form.
- Exam point: when \( a \ne 1 \), take \( a \) out of the \( x^2 \) and \( x \) terms first, and remember to multiply the correction term by \( a \).
The case: what ticket price brings in the most money?
A community theatre has 500 seats. At the current price of $20 it sells 400 tickets a night, so takings are $8,000. The manager has run a short survey and estimates that every $1 rise in price would lose about 10 sales, and every $1 cut would gain about 10. (The theatre and its numbers are illustrative.) Should the price go up, and if so, by how much?
Let \( x \) be the number of $1 rises. Then the price is \( 20 + x \) dollars, the number of tickets sold is \( 400-10x \), and the revenue is
Raising the price earns more per ticket but sells fewer tickets. Somewhere these two effects balance. The manager wants three answers: the best price, the revenue it brings, and the range of prices that would bring in at least $8,500, so that there is room to choose a round number.
The concept: making a perfect square appear
Some quadratics are perfect squares already: \( x^2-20x + 100 = (x-10)^2 \). The pattern is always the same. When you expand \( (x + p)^2 \) you get \( x^2 + 2px + p^2 \): the number in front of \( x \) is twice \( p \), and the constant is \( p^2 \).
Most quadratics are not perfect squares, but every one is a perfect square plus or minus a constant. To find it, halve the coefficient of \( x \), square the bracket, then subtract whatever extra the bracket brought in. That is completing the square.[3]
Why bother? Because a square can never be negative. In \( -10(x-10)^2 + 9000 \), the term \( -10(x-10)^2 \) is zero when \( x = 10 \) and negative for every other \( x \). So the expression can never exceed 9000, and it reaches 9000 only at \( x = 10 \). The maximum is visible without drawing a graph or using calculus.
In the first line, \( b \) is the coefficient of \( x \), \( c \) is the constant, and \( \tfrac{b}{2} \) is the number that goes inside the bracket. In the second line, \( a \) is the coefficient of \( x^2 \), \( h = -\dfrac{b}{2a} \) is the \( x \)-coordinate of the vertex, and \( k = c-\dfrac{b^2}{4a} \) is the value of the quadratic there.
Chapter 2 of Mathematics Made Simple covers completing the square, the discriminant and inequalities with worked examples.Get the textbook · $12
Why this formula fits the case
Revenue here is a quadratic in \( x \) with a negative \( x^2 \) coefficient, so its graph is an upside-down parabola with a single highest point.[4] Completing the square is designed to find exactly that point. It also gives the other two answers the manager wants. The best price comes from \( h \), the best revenue from \( k \), and the range of prices for at least $8,500 comes from solving \( R = 8500 \), which in vertex form takes one square root.
It is also the method behind the quadratic formula. Complete the square on \( ax^2 + bx + c = 0 \) and you get \( \left(x + \tfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2} \). Take square roots and you have \( x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \). So completing the square is not a separate trick; it is the reason the formula works.
When is it not the right tool? If the relationship is not quadratic, such as sales that fall by a percentage rather than a fixed number, completing the square does not apply, and calculus is the general method. And if you only need the roots of a quadratic that factorises neatly, factorising is quicker.
Setting it up
| Quantity | Meaning | In the case |
|---|---|---|
| \( x \) | Variable | Number of $1 price rises (negative means a cut) |
| \( a \) | Coefficient of \( x^2 \) | \( -10 \) |
| \( b \) | Coefficient of \( x \) | \( 200 \) |
| \( c \) | Constant | \( 8000 \), today’s revenue |
| Domain | Sensible values of \( x \) | \( -10 \le x \le 40 \): no more than 500 seats, and no negative sales |
Units matter for the reading of the answer: \( x \) is in dollars of price change and \( R \) is in dollars of revenue. Check the model at \( x = 0 \): \( R = 8000 \), today’s takings, as it should be.
Solving it
Write \( R = 8000 + 200x-10x^2 \) in vertex form, and hence find the price that maximises revenue, the number of tickets sold, and the maximum revenue.
Raising the price from $20 to $30 loses 100 sales but lifts takings by $1,000 a night.
The manager would like a range rather than a single price. The vertex form makes that easy too.
Find the range of ticket prices that brings in at least $8,500.
Any whole-dollar price from $23 to $37 keeps nightly takings at $8,500 or more.
Completing the square is also how you read a circle’s equation, which is a standard exam question.
On a map grid in kilometres, the edge of a phone mast’s coverage area is \( x^2 + y^2-6x + 8y-11 = 0 \). Find the position of the mast (the centre) and the range (the radius).
The mast stands at grid point \( (3, -4) \) and covers everything within 6 km.
Make completing the square automatic with the Practice Questions book: 1,565 questions, all fully worked.Get 1,565 questions · $9
What the answer tells us
In the case’s terms. The best price is $30, where 300 tickets bring in $9,000, compared with $8,000 now. The theatre would also have 100 fewer people to serve each night, so if costs depend on audience size, the gain in profit is even larger than the gain in revenue.
The top is flat. The vertex form shows how much a near-miss costs. Pricing at $28 or $32 instead of $30 means \( x-10 = \pm 2 \), so revenue falls by only \( 10 \times 2^2 = \$40 \). The cost grows with the square of the distance from the best price, so small errors hardly matter but large ones do: being $10 out costs $1,000.
Is it reasonable? At the best price, a $1 rise earns $1 more on each of 300 tickets, which is +$300, and loses 10 tickets at $30, which is −$300. The two effects cancel, which is exactly what should happen at a maximum.
Sensitivity. The answer depends on the estimate of 10 lost sales per $1. If it were 12, revenue would be \( (20 + x)(400-12x) \) \( = -12\left(x-\tfrac{20}{3}\right)^2 + 8533.33 \), so the best price would fall to about $26.67 and the best revenue to about $8,533. Survey quality matters.
For the full toolkit of quadratics, from factorising to the discriminant and inequalities, the Mathematics Made Simple textbook covers it in Chapter 2, and Appendix B shows the economics uses. For practice, the Mathematics Made Simple practice book has graded quadratics questions with full solutions.
Master quadratics, functions and calculus with both Mathematics Made Simple books for $16.Both books · $16
Practice questions
Show each completed square in full, then check by expanding.
EasyQ1. Write \( x^2 + 8x + 3 \) in the form \( (x + p)^2 + q \) and state the minimum value and where it occurs. Show worked answer
So \( x^2 + 8x + 3 = (x + 4)^2-13 \).
Minimum value \( -13 \), at \( x = -4 \).
EasyQ2. Solve \( x^2-6x + 2 = 0 \) by completing the square, giving exact answers and answers to 3 d.p. Show worked answer
\( x = 3 \pm \sqrt{7} \), that is \( x \approx 5.646 \) or \( x \approx 0.354 \).
MediumQ3. Express \( 2x^2 + 12x + 7 \) in the form \( a(x + b)^2 + c \), and state the coordinates of the vertex of \( y = 2x^2 + 12x + 7 \). Show worked answer
\( = 2(x + 3)^2-18 + 7 \) \( = 2(x + 3)^2-11 \).
Vertex \( (-3, -11) \), a minimum since \( a = 2 > 0 \).
MediumQ4. Find the centre and radius of the circle \( x^2 + y^2 + 10x-4y + 4 = 0 \). Show worked answer
\( (x + 5)^2 + (y-2)^2 = 25 \).
Centre \( (-5, 2) \), radius 5.
ExamQ5. A café sells 120 cakes a day at $4 each. For every 25-cent rise in price it expects to sell 6 fewer cakes. Let \( x \) be the number of 25-cent rises. (a) Show that the daily revenue is \( R = 480 + 6x-1.5x^2 \). [2] (b) Express \( R \) in the form \( k-a(x-h)^2 \). [3] (c) State the maximum revenue and the price that achieves it. [2] (d) Find the range of prices for which revenue is at least $480. [2] Show worked answer
(b) \( R = -1.5(x^2-4x) + 480 \) [1] \( = -1.5\left[(x-2)^2-4\right] + 480 \) [1] \( = 486-1.5(x-2)^2 \). [1]
(c) Maximum $486 [1] when \( x = 2 \), a price of \( 4 + 0.5 = \$4.50 \) (108 cakes). [1]
(d) \( 486-1.5(x-2)^2 \) \( \ge 480 \) \( \Rightarrow (x-2)^2 \) \( \le 4 \) \( \Rightarrow 0 \) \( \le x \) \( \le 4 \) [1], so prices from $4.00 to $5.00. [1]
Frequently asked questions
What is completing the square used for?
Three main things: finding the maximum or minimum of a quadratic and where it happens, solving quadratic equations (it is how the quadratic formula is derived), and rewriting circle equations to read off the centre and radius. It also appears later in integration and statistics.
How do you complete the square when the coefficient of \( x^2 \) is not 1?
Take the coefficient \( a \) out of the \( x^2 \) and \( x \) terms only, complete the square inside the bracket, and then multiply the subtracted constant by \( a \) as you remove the outer bracket. Q3 above shows every step.
Is completing the square better than the quadratic formula?
For solving an equation, the formula is usually quicker. For anything about the shape of the graph (vertex, maximum, minimum, range, line of symmetry), the completed square form is better, because it shows the answers directly.
Can I find a maximum revenue with calculus instead?
Yes. Differentiating \( R = 8000 + 200x-10x^2 \) gives \( R’ = 200-20x \), which is zero at \( x = 10 \), the same answer. Completing the square gets there without calculus, which is why it appears early in the course.
References
- Cambridge International Education. Cambridge International AS & A Level Mathematics 9709: syllabus for examination in 2026 and 2027, Pure Mathematics 1, topics 1.1 Quadratics and 1.3 Coordinate geometry. cambridgeinternational.org
- International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide (first assessment 2021), topic 2.6. Cardiff: IBO.
- Abramson, J. et al. (2021). Algebra and Trigonometry 2e, Section 5.1 “Quadratic Functions”. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
- Sydsæter, K., Hammond, P., Strøm, A. and Carvajal, A. (2016). Essential Mathematics for Economic Analysis, 5th edn, sections on quadratic functions. Harlow: Pearson.
