How to complete the square
The four-step method for completing the square, one fully worked example, and the three slips that lose marks in GCSE exams.
What the examiner is testing
Completing the square asks whether you can rewrite a quadratic so its structure becomes visible. The marks are for the rearrangement, not the final number, which is why a correct answer with no working still loses most of them.
The method
For any quadratic \( ax^2 + bx + c \):
- If \( a \neq 1 \), factor \( a \) out of the first two terms only.
- Halve the coefficient of \( x \). Call it \( p \).
- Write \( (x + p)^2 \). This introduces an unwanted \( p^2 \).
- Subtract \( p^2 \) to cancel it, then tidy the constant.
The whole method is step 4. Everything else is bookkeeping.
Worked example
Write \( 3x^2 - 12x + 5 \) in the form \( a(x + p)^2 + q \).
Step 1 — factor out the 3, from the first two terms only:
$$ 3x^2 - 12x + 5 = 3(x^2 - 4x) + 5 $$
The 5 stays outside. This is where most marks are lost.
Step 2 — halve the coefficient of \( x \) inside the bracket.
Half of \( -4 \) is \( -2 \), so \( p = -2 \).
Step 3 — write the square and subtract what it added.
\( (x - 2)^2 = x^2 - 4x + 4 \), so it introduces an extra 4:
$$ x^2 - 4x = (x - 2)^2 - 4 $$
Step 4 — substitute back and multiply through by 3.
$$ 3\left[(x - 2)^2 - 4\right] + 5 = 3(x - 2)^2 - 12 + 5 = 3(x - 2)^2 - 7 $$
Answer: \( 3(x - 2)^2 - 7 \), so \( a = 3 \), \( p = -2 \), \( q = -7 \).
Reading it off: the turning point is \( (2, -7) \) — a minimum, because \( a > 0 \). Since the minimum value is negative, the curve does cross the x-axis twice.
Check: expand it back. \( 3(x^2 - 4x + 4) - 7 = 3x^2 - 12x + 12 - 7 = 3x^2 - 12x + 5 \). It matches. Expanding back takes fifteen seconds and catches almost every arithmetic slip.
The three mistakes that lose marks
1. Multiplying the constant by \( a \) as well. In the example, writing \( 3(x^2 - 4x) + 15 \) instead of \( +5 \). Factor out of the first two terms only.
2. Forgetting to multiply the \( -p^2 \) back by \( a \). Writing \( 3(x-2)^2 - 4 + 5 \) gives \( 3(x-2)^2 + 1 \), which is wrong. The \( -4 \) is inside the bracket, so it becomes \( -12 \).
3. Sign errors reading off the turning point. From \( a(x + p)^2 + q \) the x-coordinate is \( -p \). With \( (x - 2)^2 \) you have \( p = -2 \), so the turning point is at \( x = +2 \).
30-second recap
Factor \( a \) from the first two terms. Halve the x-coefficient. Square the bracket, subtract the square you just created, multiply back by \( a \), tidy. Then expand it again to check. Turning point at \( (-p, q) \), minimum if \( a > 0 \).