Algebra · GCSE Maths
Completing the square
GCSE Maths completing the square: rewrite x² + bx + c as (x + p)² + q, read the turning point, and handle a ≠ 1 by factoring a out first.
x² + bx becomes (x + b/2)² − (b/2)². The turning point is (−p, q) from (x + p)² + q. If a ≠ 1, factor a out of the x terms before you complete the square.
The important bits
What you need to know
- 1
Completing the square writes a quadratic as a squared bracket plus a constant. For x² + 6x + 5: (x + 3)² − 9 + 5 = (x + 3)² − 4.
- 2
The number in the bracket is half the coefficient of x. Half of 6 is 3; half of −10 is −5. Square that number and subtract it to compensate.
- 3
From y = (x + p)² + q the turning point is (−p, q). y = (x + 3)² − 4 turns at (−3, −4). The square is never negative, so that y-value is a minimum if a > 0.
- 4
If a is negative, the completed square shows a maximum: y = −(x − 1)² + 7 turns at (1, 7) and opens downwards.
- 5
When a ≠ 1, factor a from the x² and x terms first: 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5.
- 6
To solve (x + 3)² − 4 = 0, add 4, square-root both sides: x + 3 = ±2, so x = −1 or x = −5. Completing the square is a solving method as well as a graph method.
- 7
The axis of symmetry is x = −p, which is also x = −b / (2a). Completing the square makes that line visible without the formula.
- 8
Expand your completed square immediately as a check. (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5. If you do not recover the original, p or q is wrong.
Quotations worth analysing
Short evidence. Real method.
“(x + p)² + q”
The turning point is (−p, q). Students who read (p, q) from the bracket have the x-coordinate with the wrong sign, a one-mark disaster on an otherwise perfect rewrite.
“The minimum value is q”
A square is at least 0, so (x + p)² + q is at least q. That minimum occurs when the square is 0, at x = −p. If a > 1, the minimum is the completed constant after scaling.
“Factor out a before completing the square”
You complete the square on x² + (b/a)x, then multiply the square and the compensating constant by a. Skipping the factor produces a bracket that does not match the original quadratic.
Go deeper
Half, square, subtract, then tidy q
Start with x² + 10x + 7. Half of 10 is 5, and 5² is 25. So x² + 10x = (x + 5)² − 25. Add the 7: (x + 5)² − 25 + 7 = (x + 5)² − 18. Turning point (−5, −18). To solve (x + 5)² − 18 = 0, (x + 5)² = 18, x + 5 = ±√18 = ±3√2, x = −5 ± 3√2. The same steps with a minus in the middle: x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13, turning point (4, −13). Half of −8 is −4, and the bracket is (x − 4). Students who always write (x + something) then lose the sign. Check by expanding; it takes ten seconds and catches a 25 that should have been 16.
Go deeper
When a is not 1
2x² + 12x + 5 looks hostile until you factor 2 from the terms that contain x: 2(x² + 6x) + 5. Now complete the square inside: x² + 6x = (x + 3)² − 9. So 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13. Turning point (−3, −13). The 2 stretches the U but does not move the x-coordinate of the vertex. Minimum value is −13, not −9, because of the stretch and the leftover constant. If a is −1 or −2, factor the negative as well: −x² + 6x + 1 = −(x² − 6x) + 1 = −[(x − 3)² − 9] + 1 = −(x − 3)² + 9 + 1 = −(x − 3)² + 10. Maximum 10 at x = 3. The sign of a is the difference between a floor and a ceiling on the graph.
Go deeper
Turning point questions are this method in costume
Higher papers ask for the coordinates of the turning point, the minimum value, or the line of symmetry, often without the words “complete the square”. That is still this page. You can also use x = −b / (2a) then substitute into y, which is faster numerically, but completing the square shows why that x is special: it is where the square vanishes. Sketch a tiny U or n with the vertex labelled; examiners like to see that you know whether it is a min or a max. If a later part asks you to solve f(x) = 0, stay in completed-square form and square-root; do not expand back to x² + bx + c and start again with the formula unless the numbers are nicer that way.
See the idea in action
Write 2x² + 8x + 3 in the form a(x + p)² + q and state the turning point. Step 1: Factor 2 from the x terms: 2(x² + 4x) + 3. Step 2: Half of 4 is 2, and 2² = 4, so x² + 4x = (x + 2)² − 4. Step 3: 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5. Step 4: a = 2, p = 2, q = −5. Turning point is (−2, −5). Check by expanding: 2(x² + 4x + 4) − 5 = 2x² + 8x + 8 − 5 = 2x² + 8x + 3. The 2 > 0 so this is a minimum value of −5.
Exam technique
Turn knowledge into marks
The x-coordinate of the turning point is the opposite sign of the number in the bracket. From (x + 2)² the vertex is at x = −2, never at x = 2.
Common mistakes
Do not give these marks away
- 01
Reading the turning point as (p, q) instead of (−p, q) from (x + p)² + q.
- 02
Completing the square without factoring out a when a ≠ 1, so the expansion does not match the original quadratic.
- 03
Forgetting to subtract (b/2)², leaving (x + 3)² + 5 when it should be (x + 3)² − 9 + 5.
x² + 6x + 5 written as (x + p)² + q is
A(x + 3)² − 4
B(x + 3)² + 5
C(x + 6)² − 31
D(x − 3)² − 4
Show the answer
(x + 3)² − 4. Half of 6 is 3, so (x + 3)² − 9 + 5 = (x + 3)² − 4. (x + 3)² + 5 forgot to subtract 9. (x − 3)² − 4 has the wrong sign in the bracket. (x + 6)² used b instead of b/2.
Quick questions
If this is the bit you searched
How do you complete the square GCSE?
Take half the coefficient of x, write (x + b/2)², subtract (b/2)², then absorb the constant. If a ≠ 1, factor a from the x² and x terms first.
How do you find the turning point from completed square form?
From y = a(x + p)² + q the turning point is (−p, q). If a > 0 it is a minimum; if a < 0 it is a maximum.
When should I complete the square instead of using the formula?
When the question asks for a turning point, a minimum or maximum value, or the form (x + p)² + q. Use the formula when you only need the roots and the numbers are ugly.
What if the coefficient of x is odd, such as x² + 5x?
Half of 5 is 5/2. Write (x + 5/2)² − 25/4. Fractions are normal; do not round them. Keep improper fractions rather than mixed numbers inside the algebra.