Complete the square when the leading coefficient is not one
Factor the leading coefficient correctly and connect the resulting square to graph shape and optimisation.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Complete a square when the coefficient of x² is one.
- Distribute a constant over a bracket, including a negative constant.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which bracket equals 3x² + 12x?
What you will learn
- Complete the square for ax² + bx + c with a ≠ 0.
- Use the sign of a to distinguish a minimum from a maximum.
Factor before halving
Begin by taking a out of the x² and x terms: a[x² + (b/a)x] + c. Inside this bracket the coefficient to halve is b/a, not b. Complete that inner square and remember that its correction is also multiplied by a.
For 2x² − 12x + 13, first write 2(x² − 6x) + 13. The inner square is (x − 3)² − 9, so the complete expression is 2(x − 3)² − 18 + 13 = 2(x − 3)² − 5. A common error is subtracting only 9 outside the factor of 2.
Interpret a(h − x)² just as carefully
In y = a(x − h)² + k, a > 0 opens upward and gives a minimum k at x = h; a < 0 opens downward and gives a maximum k there. These statements assume the vertex lies in the permitted domain. The magnitude of a changes the vertical scale; the symmetry line remains x = h.
For −2x² + 8x − 3, factor −2 from the first two terms to get −2(x² − 4x) − 3. Completing the square gives −2(x − 2)² + 5. The vertex is (2, 5), and y cannot exceed 5 over all real x. Expansion is an effective check on every sign and constant.
Pause and explain
What is the maximum of y = 5 − 2(x − 2)² over all real x?
Worked example
Complete the square for y = 2x² − 12x + 13. State the vertex and its type.
Show the worked solution
- Factor 2 from the quadratic and linear terms: 2(x² − 6x) + 13.
- Complete the inner square: 2[(x − 3)² − 9] + 13.
- Simplify to 2(x − 3)² − 5; the positive coefficient makes its vertex a minimum.
Answer y = 2(x − 3)² − 5; minimum at (3, −5).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Complete the square for 3x² + 12x + 7 and give its minimum.
Give me a hint
Take 3 out of the first two terms; complete x² + 4x.
Compare my reasoning
- Write 3(x² + 4x) + 7.
- Replace the inner expression by (x + 2)² − 4, giving 3(x + 2)² − 12 + 7.
- Simplify to 3(x + 2)² − 5, whose minimum occurs when x = −2.
3(x + 2)² − 5; minimum −5 at x = −2.
Look for these in your work
- Multiply the correction −4 by 3.
- Use the positive square multiplier to justify the minimum.
Complete the square for −2x² + 8x − 3 and state its greatest value over real x.
Give me a hint
Take out −2 and keep that sign when distributing the correction.
Compare my reasoning
- Write −2(x² − 4x) − 3.
- Complete the inner square: −2[(x − 2)² − 4] − 3.
- Simplify to 5 − 2(x − 2)²; its maximum is 5 at x = 2.
5 − 2(x − 2)²; greatest value 5 at x = 2.
Look for these in your work
- Handle the negative multiplier on both inner terms.
- Explain why the squared term can only reduce the value from 5.
A rectangle has perimeter 20 cm. Its side lengths are x and 10 − x cm, with 0 < x < 10. Use completing the square to find the greatest area.
Give me a hint
Express the area as x(10 − x) before rewriting it.
Compare my reasoning
- The area is A = 10x − x² = −(x² − 10x).
- Complete the square: A = 25 − (x − 5)².
- The permitted value x = 5 gives area 25 and both sides 5 cm.
Greatest area 25 cm², for a 5 cm by 5 cm square.
Look for these in your work
- Use the perimeter to obtain the correct second side.
- Check the optimum satisfies both positive-side constraints.
Common mistakes
- After factoring a, halve b/a rather than b.
- Multiply the square's correction by a before combining constants.
- A negative leading coefficient gives a maximum, not a minimum.
What can you explain now?
Complete the square for 4x² − 20x + 21 and give the vertex.
Compare with the explanation
4(x − 5/2)² − 4; vertex (5/2, −4), a minimum.
Factor 4, then subtract 4 × 25/4 = 25 from the original constant 21.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain why the correction outside a completed square changes when a ≠ 1. Reconstruct the rectangle argument without differentiating.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.