Completing the Square to Find the Minimum

Completing the square is worth learning cold because it answers three question types at once: the minimum (or maximum) point, the line of symmetry, and the completed-square form itself. When the leading coefficient is not $1$, the only extra care needed is factoring it out first.

(i) Express $2x^2 - 12x + 5$ in the form $a(x + b)^2 + c$, where $a$, $b$ and $c$ are constants. [3] (ii) Hence state the minimum value of $2x^2 - 12x + 5$ and the value of $x$ at which it occurs. [2]

The working

(i) Factor the $2$ out of the $x$-terms only: $2x^2 - 12x + 5 = 2(x^2 - 6x) + 5 \quad \text{(M1)}$

Complete the square inside the bracket. Half of $-6$ is $-3$, so $x^2 - 6x = (x - 3)^2 - 9$: $= 2\big[(x - 3)^2 - 9\big] + 5 \quad \text{(M1)}$

Multiply the $2$ back in and combine constants ($-18 + 5 = -13$): $= 2(x - 3)^2 - 13 \quad \text{(A1)}$

So $a = 2$, $b = -3$, $c = -13$.

(ii) A squared term is never negative, so $2(x - 3)^2 \ge 0$, and the whole expression is smallest when the bracket is zero: $\text{minimum value} = -13, \quad \text{occurring at } x = 3 \quad \text{(A1, A1)}$

Where the marks are won and lost

• The middle-step accuracy mark depends on multiplying the inside constant by the outside factor. $2[(x-3)^2 - 9]$ becomes $2(x-3)^2 - 18$, not $2(x-3)^2 - 9$.
• The minimum value is $c = -13$; the $x$-value is whatever makes the bracket zero, here $x = 3$ (note: the bracket is $(x - 3)$, so $x = +3$, not $-3$).
• “Hence” in (ii) means use part (i), do not restart with calculus or a table of values. The completed square already contains the answer.

Common mistakes

• Forgetting to multiply $-9$ by $2$, giving a minimum of $-4$ instead of $-13$.
• Reading the turning point as $x = -3$ from the $(x - 3)$ bracket.
• Halving $-12$ (the original coefficient) instead of $-6$ after factoring out the $2$.

Full method: Completing the Square notes. See also Maximum/Minimum & the Vertex and the Quadratic Functions pillar.