IGCSE · Practice questions · Completing the square: maximum and minimum values

Square it, then read it.

Six original Cambridge-style questions on completing the square: minimum and maximum values and where they occur, exact fractional answers, finding constants from a turning point, and an area problem solved without calculus.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Take out \(a\), halve, square, adjust, read.

Take out \(a\) from the \(x^{2}\) and \(x\) terms only (with its sign if \(a\) is negative).
Halve the coefficient of \(x\) inside the bracket: \(x^{2} + bx = \left(x + \frac{b}{2}\right)^{2} - \left(\frac{b}{2}\right)^{2}\).
Multiply back the subtracted square by \(a\), then combine with the constant.
Read \(a(x + p)^{2} + q\): \(a > 0\) gives a minimum \(q\), \(a < 0\) a maximum \(q\), at \(x = -p\).
Keep it exact: fractions such as \(-\frac{17}{4}\) stay as fractions on Paper 1.

Recall

[2 marks]

Express \(x^{2} - 12x + 40\) in the form \((x + p)^{2} + q\). Hence state the minimum value of \(x^{2} - 12x + 40\) and the value of \(x\) at which it occurs.

Half of \(-12\) is \(-6\): \(x^{2} - 12x + 40 = (x - 6)^{2} - 36 + 40 = (x - 6)^{2} + 4\). ✓
Minimum value 4 when \(x = 6\). ✓

Calculation

[3 marks]

Find the minimum value of \(x^{2} + 5x + 2\) and the value of \(x\) at which it occurs. Give your answers as exact fractions.

\(x^{2} + 5x = \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4}\). ✓
\(x^{2} + 5x + 2 = \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4} + \dfrac{8}{4} = \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{17}{4}\). ✓
Minimum value \(-\dfrac{17}{4}\) when \(x = -\dfrac{5}{2}\). ✓

Calculation

[3 marks]

Express \(4x^{2} + 20x + 19\) in the form \(a(x + b)^{2} + c\), where \(a\), \(b\) and \(c\) are constants. Hence state the minimum value of \(4x^{2} + 20x + 19\).

\(4(x^{2} + 5x) + 19 = 4\left[\left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4}\right] + 19\). ✓
\(= 4\left(x + \dfrac{5}{2}\right)^{2} - 25 + 19 = 4\left(x + \dfrac{5}{2}\right)^{2} - 6\). ✓
Minimum value \(-6\), when \(x = -\dfrac{5}{2}\). ✓

Calculation

[4 marks]

The function f is defined by \(f(x) = 7 - 12x - 3x^{2}\) for \(x \in \mathbb{R}\).

(a) Express \(f(x)\) in the form \(a - b(x + c)^{2}\).

(b) Write down the maximum value of \(f(x)\), the value of \(x\) at which it occurs, and the range of f.

(a) \(-3x^{2} - 12x = -3(x^{2} + 4x) = -3\left[(x + 2)^{2} - 4\right] = -3(x + 2)^{2} + 12\). ✓
So \(f(x) = 19 - 3(x + 2)^{2}\): \(a = 19,\ b = 3,\ c = 2\). ✓
(b) \(3(x + 2)^{2} \ge 0\) is subtracted from 19, so the maximum value is 19, when \(x = -2\). ✓
Range: \(f(x) \le 19\). ✓

Analysis

[4 marks]

The expression \(2x^{2} + px + q\) has a minimum value of 5, which occurs when \(x = 3\). Find the values of the constants \(p\) and \(q\).

The coefficient of \(x^{2}\) is 2 and the minimum is 5 at \(x = 3\), so \(2x^{2} + px + q = 2(x - 3)^{2} + 5\). ✓
Expand: \(2(x^{2} - 6x + 9) + 5 = 2x^{2} - 12x + 18 + 5\). ✓
\(= 2x^{2} - 12x + 23\). ✓
Comparing coefficients: \(p = -12,\ q = 23\). ✓

Analysis

[5 marks]

A rectangular pen is built against a long straight wall, so fencing is needed on three sides only. The two sides at right angles to the wall each have length \(x\) m, and exactly 60 m of fencing is used. The area of the pen is \(A\) m².

(a) Show that \(A = 60x - 2x^{2}\).

(b) By completing the square, find the greatest possible area of the pen and the value of \(x\) that gives it.

(a) The side parallel to the wall is \(60 - 2x\) m, so \(A = x(60 - 2x) = 60x - 2x^{2}\). ✓
(b) \(A = -2(x^{2} - 30x) = -2\left[(x - 15)^{2} - 225\right] = 450 - 2(x - 15)^{2}\). ✓
\(2(x - 15)^{2} \ge 0\), so the greatest area is 450 m², when \(x = 15\) (the pen is 15 m by 30 m). ✓
(c) \(A = 450 - 2(x - 15)^{2} \le 450\) for every \(x\). ✓
Since \(500 > 450\), an area of 500 m² cannot be made with 60 m of fencing. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.