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
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 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.
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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.