The shape of the lesson
By the end of the lesson, learners can
- Half of b, then square: In the lab set \(b\) to 6 and press Make the square.
- When a is not 1: Take \(a\) out as a factor of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out the number you subtracted.
- Solving and turning points: Once a quadratic is in completed square form, the unknown appears only once, so you can solve by undoing: add, square root (both signs), subtract.
- \(a \ne 1\): take \(a\) out of the \(x^{2}\) and \(x\) terms first.
- Halve: \(x^{2} + bx = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}\).
- Tidy: combine the constants; multiply back by \(a\) if you took it out.
- Solve: isolate the square, square root with \(\pm\), then rearrange.
- Turning point: \(y = a(x + p)^{2} + q\) turns at \((-p,\ q)\).
One picture every idea returns to
Halve the coefficient of \(x\): \(x^{2} + bx = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}\). If \(a \ne 1\), take \(a\) out of the \(x^{2}\) and \(x\) terms first. The graph of \(y = (x + p)^{2} + q\) has its minimum point at \((-p,\ q)\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Make a perfect square.. Extended. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Half of b, then square | In the lab set \(b\) to 6 and press Make the square. Project tab 4 of the lab (Algebra lab, completing the square) and let a student drive it while the class predicts. Model the worked example Positive b on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: When a is not 1 | Take \(a\) out as a factor of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out the number you subtracted. Model the worked example Factor out a on the board, one line at a time. Your turn: Write \(2x^{2} - 8x + 3\) in the form \(a(x + p)^{2} + q\). | Pairs |
| 21 to 30 min | Teach and check: Solving and turning points | Once a quadratic is in completed square form, the unknown appears only once, so you can solve by undoing: add, square root (both signs), subtract. Model the worked example Exact solutions on the board, one line at a time. Your turn: Solve \(x^{2} + 8x + 3 = 0\) by completing the square, giving exact answers and then answers correct to 2 decimal places. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Solving and turning points, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The number you subtract is \(\left(\dfrac{b}{2}\right)^{2}\), and it is always subtracted, even when \(b\) is negative: \(x^{2} - 10x = (x - 5)^{2} - 25\), not \(+ 25\). | Return to Half of b, then square: ask a pair to state the correct version and show it on the lab before moving on. |
| In \(2[(x + 3)^{2} - 9]\) the 2 multiplies the \(-9\) too, giving \(-18\). | Subtracting only 9 gives the wrong constant \(-2\). |
| Square rooting gives two values: \(x - 3 = +\sqrt{5}\) or \(-\sqrt{5}\). | And the turning point of \((x - 2)^{2} - 3\) is at \(x = +2\), the opposite sign to the number in the bracket. |
Completing the square: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Write in the form \((x + p)^{2} + q\): (a) \(x^{2} + 8x + 3\) (b) \(x^{2} - 2x - 5\).
Write \(x^{2} + 5x - 2\) in the form \((x + p)^{2} + q\).
Write \(3x^{2} - 12x + 5\) in the form \(a(x + p)^{2} + q\).
By completing the square, solve \(x^{2} + 8x + 3 = 0\).
Give your answers (a) in exact form (b) correct to 2 decimal places.
(a) Write \(5 + 4x - x^{2}\) in the form \(r - (x + p)^{2}\).
(b) Hence write down the maximum value of \(5 + 4x - x^{2}\) and the value of \(x\) where it occurs.
(a) Write \(x^{2} - 4x + 7\) in the form \((x + p)^{2} + q\).
(b) Explain why the equation \(x^{2} - 4x + 7 = 0\) has no solutions.
(c) Find the greatest possible value of \(\dfrac{12}{x^{2} - 4x + 7}\).
Answer key and marking notes
- (a) \((x + 4)^{2} - 16 + 3 = (x + 4)^{2} - 13\)
- (b) \((x - 1)^{2} - 1 - 5 = (x - 1)^{2} - 6\)
- \(\left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4} - 2\)
- \(= \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{33}{4}\)
- \(3(x^{2} - 4x) + 5\)
- \(= 3[(x - 2)^{2} - 4] + 5\)
- \(= 3(x - 2)^{2} - 7\)
- (a) \((x + 4)^{2} - 13 = 0\), so \((x + 4)^{2} = 13\)
- \(x + 4 = \pm\sqrt{13}\), \(x = -4 \pm \sqrt{13}\)
- (b) \(x = -0.39\) or \(x = -7.61\)
- (a) \(-(x^{2} - 4x) + 5 = -[(x - 2)^{2} - 4] + 5 = 9 - (x - 2)^{2}\)
- (b) \((x - 2)^{2} \ge 0\), so the expression is largest when \((x - 2)^{2} = 0\)
- Maximum value 9, when \(x = 2\)
- (a) \((x - 2)^{2} - 4 + 7 = (x - 2)^{2} + 3\)
- (b) \((x - 2)^{2} \ge 0\), so \((x - 2)^{2} + 3 \ge 3\). It can never equal 0.
- (c) The fraction is greatest when the denominator is least
- The least value of \(x^{2} - 4x + 7\) is 3 (when \(x = 2\))
- Greatest value \(= \dfrac{12}{3} = 4\)
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Write \(x^{2} + 10x + 18\) in the form \((x + p)^{2} + q\).
- A\((x + 5)^{2} - 7\)
- B\((x + 5)^{2} + 43\)
- C\((x + 10)^{2} - 82\)
- D\((x + 5)^{2} + 18\)
Write \(x^{2} - 3x + 1\) in the form \((x + p)^{2} + q\).
- A\(\left(x - \dfrac{3}{2}\right)^{2} - \dfrac{5}{4}\)
- B\(\left(x - \dfrac{3}{2}\right)^{2} + \dfrac{13}{4}\)
- C\(\left(x + \dfrac{3}{2}\right)^{2} - \dfrac{5}{4}\)
- D\((x - 3)^{2} - 8\)
Write \(3x^{2} - 12x + 5\) in the form \(a(x + p)^{2} + q\).
- A\(3(x - 2)^{2} - 7\)
- B\(3(x - 2)^{2} + 1\)
- C\(3(x + 2)^{2} - 7\)
- D\(3(x - 6)^{2} - 103\)
Solve \(x^{2} + 4x - 9 = 0\) by completing the square. Give your answers in exact form.
- A\(x = -2 \pm \sqrt{13}\)
- B\(x = 2 \pm \sqrt{13}\)
- C\(x = -2 \pm \sqrt{5}\)
- D\(x = -4 \pm \sqrt{13}\)
Find the coordinates of the turning point of the graph of \(y = x^{2} + 8x + 11\).
- A\((-4,\ -5)\)
- B\((4,\ -5)\)
- C\((-4,\ 27)\)
- D\((-8,\ 11)\)
By writing \(7 + 6x - x^{2}\) in the form \(r - (x + p)^{2}\), find the maximum value of \(7 + 6x - x^{2}\).
- A\(16\)
- B\(7\)
- C\(-2\)
- D\(3\)
Quiz key
\(x^{2} + 10x = (x + 5)^{2} - 25\), so \(x^{2} + 10x + 18 = (x + 5)^{2} - 25 + 18 = (x + 5)^{2} - 7\). \((x + 5)^{2} + 43\) adds the 25 instead of subtracting it. \((x + 10)^{2}\) forgets to halve the coefficient of \(x\).
\(x^{2} - 3x = \left(x - \dfrac{3}{2}\right)^{2} - \dfrac{9}{4}\). Then \(-\dfrac{9}{4} + 1 = -\dfrac{5}{4}\). \(+\dfrac{13}{4}\) adds the \(\dfrac{9}{4}\) instead of subtracting it. \(\left(x + \dfrac{3}{2}\right)^{2}\) loses the minus sign of \(-3x\). \((x - 3)^{2} - 8\) does not halve the 3.
\(3(x^{2} - 4x) + 5 = 3[(x - 2)^{2} - 4] + 5 = 3(x - 2)^{2} - 12 + 5 = 3(x - 2)^{2} - 7\). \(3(x - 2)^{2} + 1\) subtracts 4 instead of \(3 \times 4 = 12\). \(3(x - 6)^{2}\) halves 12 without first taking out the 3.
\((x + 2)^{2} - 4 - 9 = 0\), so \((x + 2)^{2} = 13\), \(x + 2 = \pm\sqrt{13}\), \(x = -2 \pm \sqrt{13}\). \(-2 \pm \sqrt{5}\) adds the 4 instead of subtracting it. \(2 \pm \sqrt{13}\) gets the sign wrong when moving the 2 across.
\(x^{2} + 8x + 11 = (x + 4)^{2} - 16 + 11 = (x + 4)^{2} - 5\). The bracket is smallest (zero) when \(x = -4\), so the turning point is \((-4, -5)\). \((4, -5)\) uses \(+p\) instead of \(-p\). \((-4, 27)\) adds the 16. \((-8, 11)\) does not halve the 8.
\(-(x^{2} - 6x) + 7 = -[(x - 3)^{2} - 9] + 7 = 16 - (x - 3)^{2}\). A square is never negative, so the expression is largest when \((x - 3)^{2} = 0\), giving 16. \(-2\) forgets that the minus outside turns \(-9\) into \(+9\). 3 is the value of \(x\) where the maximum happens, not the maximum value. 7 is just the value when \(x = 0\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.