To complete the square on ax² + bx + c, factorise a out of the x terms, add and subtract the square of half the new coefficient, then multiply the subtracted term back through by a. The result is a(x + p)² + q.
This appears in the first lessons of quadratic structure and discriminants and returns in turning points, circles and integration later.
How does the method work when a is not 1?
The idea is to make a perfect square from the x terms only. When a = 1 that is direct. When a ≠ 1, pull a out of the x² and x terms first so the bracket starts with x² again.
- Factorise a from the x² and x terms only. Leave c outside.
- Halve the coefficient of x inside the bracket and square it.
- Add and subtract that square inside the bracket.
- Write the perfect square, leaving the subtracted term in the bracket.
- Multiply the leftover term by a to bring it out of the bracket, then combine it with c.
Worked example
Write 2x² − 12x + 7 in the form a(x + p)² + q, and state the minimum value.
Step 1, factorise a from the x terms: 2x² − 12x + 7 = 2(x² − 6x) + 7.
Step 2, half of −6 is −3, and (−3)² = 9.
Step 3, add and subtract 9 inside the bracket: 2(x² − 6x + 9 − 9) + 7.
Step 4, perfect square: 2[(x − 3)² − 9] + 7.
Step 5, multiply the −9 by 2: 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11.
The minimum value is −11, at x = 3, because 2(x − 3)² is never negative.
Check: at x = 0, the original gives 7 and the new form gives 2 × 9 − 11 = 7. At x = 1, the original gives 2 − 12 + 7 = −3 and the new form gives 2 × 4 − 11 = −3. Both agree.
The mistake to watch for
The most common slip is to leave the −9 un-multiplied.
Mistaken answer: 2(x − 3)² − 9 + 7 = 2(x − 3)² − 2
The student kept the 2 in front of the bracket but forgot that the −9 is also inside the 2[ ].
Test it at x = 0: the mistaken form gives 2 × 9 − 2 = 16, but the original is 7. The correction is to write the outer bracket explicitly, 2[(x − 3)² − 9], and expand the 2 over both terms before combining with c.
Check yourself
Use the substitution check on each one.
1. Complete the square: x² + 8x + 3.
Show answer
Half of 8 is 4, and 4² = 16. So x² + 8x + 3 = (x + 4)² − 16 + 3 = (x + 4)² − 13. Check at x = 0: 16 − 13 = 3.
2. Write 3x² + 12x − 5 in the form a(x + p)² + q.
Show answer
3(x² + 4x) − 5 = 3[(x + 2)² − 4] − 5 = 3(x + 2)² − 12 − 5 = 3(x + 2)² − 17. Check at x = 0: 12 − 17 = −5.
3. Write 5 − 4x − x² in completed-square form and state its greatest value.
Show answer
Write it as −x² − 4x + 5 = −(x² + 4x) + 5 = −[(x + 2)² − 4] + 5 = −(x + 2)² + 9. The greatest value is 9, at x = −2. Check at x = 1: 5 − 4 − 1 = 0, and −9 + 9 = 0.
Where this leads next
Next, see how the same form reads off a minimum in connect a turning point with a minimum value, and how the discriminant tells you about roots in use the discriminant to classify intersections. The quadratic structure explorer lets you change a, b and c and watch the completed square update, and the non-calculator working trainer is good for keeping fractions exact.
If the method is clear in a lesson but breaks in a timed paper, that is the pattern our teachers look for in online one-to-one Additional Mathematics tuition.