Completing the square and locating a turning point
Rewrite a quadratic as a shifted square so its minimum or maximum becomes visible.
Extended only.
Before you begin
- Expand a squared binomial.
- Work with fractions and both square-root signs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is (x + 3)²?
What you will learn
- Complete the square with a leading coefficient of one or another non-zero value.
- Read a quadratic's symmetry and turning point.
- Solve a quadratic by reversing the square.
Build the square and compensate
The expression (x + p)² expands to x² + 2px + p². To match x² + 6x + 2, choose p = 3, then subtract the extra nine introduced by the square: (x + 3)² − 9 + 2 = (x + 3)² − 7. Adding and subtracting the same value preserves the expression.
When the middle coefficient is odd, p can be a fraction. Thus x² − 5x + 1 = (x − 5/2)² − 25/4 + 1 = (x − 5/2)² − 21/4. Keep the exact fraction rather than rounding halfway through the transformation.
Factor the leading coefficient before the square
For 2x² − 8x + 5, first write 2(x² − 4x) + 5. The bracket becomes (x − 2)² − 4, so the whole expression is 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3. The compensation inside the bracket must also be multiplied by two.
For a negative leading coefficient, the method still works. The expression −x² + 4x + 1 becomes −(x − 2)² + 5. Since a square is non-negative, subtracting it can only decrease the value from five. This explains a maximum rather than a minimum.
Read and use the new form
In y = a(x − h)² + k, the square is zero at x = h, so the turning point is (h, k) and the symmetry line is x = h. If a > 0, k is the minimum; if a < 0, k is the maximum. The sign inside the bracket must be reversed when identifying h.
To solve (x − h)² = r with r > 0, take both signs: x = h ± √r. If r = 0 there is one repeated solution. If r < 0 there are no real solutions because a real square cannot be negative. This approach gives exact surd solutions and will support sketching and differentiation later.
Pause and explain
Where is the turning point of y = (x + 4)² − 2?
Worked example
Write 2x² − 8x + 5 in completed-square form and state its minimum.
Show the worked solution
- Factor two from the x terms: 2(x² − 4x) + 5.
- Complete inside: 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3.
- The square is at least zero, so the minimum is −3 at x = 2.
Answer 2(x − 2)² − 3; minimum −3 at x = 2
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 x² + 8x + 3.
Give me a hint
Half of eight is the number inside the new bracket.
Compare my reasoning
- Use (x + 4)² = x² + 8x + 16.
- Compensate for the extra 16: (x + 4)² − 16 + 3.
- Simplify the constant to obtain (x + 4)² − 13.
(x + 4)² − 13
Look for these in your work
- Halved the linear coefficient.
- Subtracted the added square.
Solve x² − 6x + 5 = 0 by completing the square.
Give me a hint
Move the constant outside the square before taking roots.
Compare my reasoning
- Write (x − 3)² − 4 = 0.
- Then (x − 3)² = 4, so x − 3 = ±2.
- The solutions are x = 1 and x = 5; each satisfies the original equation.
x = 1 or 5
Look for these in your work
- Used both square-root signs.
- Checked the original quadratic.
A model gives height h = −t² + 6t + 2 metres for 0 ≤ t ≤ 6 seconds. Find the greatest height and when it occurs.
Give me a hint
Rewrite the expression as a negative square plus a constant.
Compare my reasoning
- h = −(t² − 6t) + 2 = −(t − 3)² + 11.
- The subtracted square is zero at t = 3, which lies in the time interval.
- The greatest height is 11 m at 3 s.
11 m at t = 3 s
Look for these in your work
- Interpreted the negative leading coefficient as a maximum.
- Checked the turning time is within the model's domain.
Common mistakes
- Forgetting to compensate for the added square.
- Failing to multiply the compensation by the leading coefficient.
- Reading the bracket's sign directly as the turning point's x-coordinate.
What can you explain now?
Write x² − 4x + 7 in completed-square form.
Compare with the explanation
(x − 2)² + 3
(x − 2)² supplies x² − 4x + 4, so three more gives the original constant seven.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Complete the square for a quadratic with an odd middle coefficient. Explain the turning point from the non-negative square, then expand to verify it.
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources