The shape of the lesson
By the end of the lesson, learners can
- Solve by factorising: In the lab set \(a = 1\), \(b = -5\), \(c = 6\) and look at where the curve crosses the \(x\)-axis: at 2 and 3.
- The quadratic formula: The formula solves every quadratic, whether it factorises or not.
- Problems that make quadratics: Form an equation, rearrange to \(= 0\), solve, then decide which answers make sense.
- Zero: rearrange to \(ax^{2} + bx + c = 0\).
- Factorise: if you can, set each bracket to zero.
- Formula: \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\); work out \(b^{2} - 4ac\) first.
- Form: exact (surds) or rounded, as asked.
- Context: reject impossible answers and answer the actual question.
One picture every idea returns to
Rearrange to \(ax^{2} + bx + c = 0\) first. If it factorises, set each bracket equal to zero. If not, use \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). The solutions are where the curve \(y = ax^{2} + bx + c\) crosses the \(x\)-axis.
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, Two answers, usually.. Extended. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Solve by factorising | In the lab set \(a = 1\), \(b = -5\), \(c = 6\) and look at where the curve crosses the \(x\)-axis: at 2 and 3. Project tab 4 of the lab (Equations lab, quadratic equations) and let a student drive it while the class predicts. Model the worked example Rearrange, then factorise on the board, one line at a time. Your turn: Solve \(3x^{2} - 10x + 8 = 0\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: The quadratic formula | The formula solves every quadratic, whether it factorises or not. Model the worked example Answers to 2 decimal places on the board, one line at a time. Your turn: Solve \(3x^{2} + 7x - 2 = 0\), giving your answers correct to 2 decimal places. | Pairs |
| 21 to 30 min | Teach and check: Problems that make quadratics | Form an equation, rearrange to \(= 0\), solve, then decide which answers make sense. Model the worked example An area on the board, one line at a time. | Rally Coach |
| 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, 23 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 Problems that make quadratics, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Dividing \(x^{2} = 7x\) by \(x\) gives only \(x = 7\) and loses \(x = 0\). | Never divide by the unknown; factorise instead. |
| When \(b\) is negative, \(-b\) is positive: for \(b = -3\), \(-b = +3\). | And \(-4ac\) with \(c = -4\) becomes \(+32\). Writing every value in brackets prevents both slips. |
| Do not stop at the solutions of the quadratic. | Reject impossible values and answer the question that was asked, which may be a perimeter or a length rather than \(x\). |
Solving quadratic equations: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
Solve (a) \(x^{2} - 5x + 6 = 0\) (b) \(x^{2} + 3x = 10\).
Solve \(3x^{2} - 10x + 8 = 0\).
Solve \(x^{2} - 6x + 4 = 0\).
(a) Give your answers in the form \(p \pm \sqrt{q}\). (b) Give your answers correct to 2 decimal places.
Solve \(3x^{2} + 7x - 2 = 0\), giving your answers correct to 2 decimal places.
Solve \(\dfrac{2}{x + 2} + \dfrac{3}{2x - 1} = 1\).
A rectangular garden measures \((x + 4)\) m by \((2x - 1)\) m. Its area is 60 m2.
(a) Show that \(2x^{2} + 7x - 64 = 0\).
(b) Solve this equation, giving your answers correct to 2 decimal places.
(c) Find the perimeter of the garden.
Answer key and marking notes
- (a) \((x - 2)(x - 3) = 0\), \(x = 2\) or \(x = 3\)
- (b) \(x^{2} + 3x - 10 = 0\), \((x + 5)(x - 2) = 0\), \(x = -5\) or \(x = 2\)
- \(3x^{2} - 6x - 4x + 8 = 0\)
- \((3x - 4)(x - 2) = 0\)
- \(x = \dfrac{4}{3}\) or \(x = 2\)
- (a) \(b^{2} - 4ac = 36 - 16 = 20\), \(x = \dfrac{6 \pm \sqrt{20}}{2} = \dfrac{6 \pm 2\sqrt{5}}{2}\)
- \(x = 3 \pm \sqrt{5}\)
- (b) \(x = 5.24\) or \(x = 0.76\)
- \(b^{2} - 4ac = 49 - 4(3)(-2) = 73\)
- \(x = \dfrac{-7 \pm \sqrt{73}}{6}\)
- \(x = 0.26\) or \(x = -2.59\)
- Multiply by \((x + 2)(2x - 1)\): \(2(2x - 1) + 3(x + 2) = (x + 2)(2x - 1)\)
- \(7x + 4 = 2x^{2} + 3x - 2\)
- \(2x^{2} - 4x - 6 = 0\), so \(x^{2} - 2x - 3 = 0\)
- \((x - 3)(x + 1) = 0\), \(x = 3\) or \(x = -1\)
- (a) \((x + 4)(2x - 1) = 2x^{2} - x + 8x - 4 = 2x^{2} + 7x - 4\)
- \(2x^{2} + 7x - 4 = 60\), so \(2x^{2} + 7x - 64 = 0\)
- (b) \(b^{2} - 4ac = 49 + 512 = 561\), \(x = \dfrac{-7 \pm \sqrt{561}}{4}\)
- \(x = 4.17\) or \(x = -7.67\)
- (c) \(x = -7.67\) gives negative lengths, so \(x = 4.171...\): sides 8.171... m and 7.342... m
- Perimeter \(= 2(8.171... + 7.342...) = 31.0\) m (3 s.f.)
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.
Solve \(x^{2} - 2x - 24 = 0\).
- A\(x = 6\) or \(x = -4\)
- B\(x = -6\) or \(x = 4\)
- C\(x = 8\) or \(x = -3\)
- D\(x = 12\) or \(x = -2\)
Solve \(3x^{2} = 12x\).
- A\(x = 0\) or \(x = 4\)
- B\(x = 4\) only
- C\(x = 0\) or \(x = -4\)
- D\(x = 2\) or \(x = -2\)
Solve \(2x^{2} - 5x - 12 = 0\).
- A\(x = 4\) or \(x = -1.5\)
- B\(x = -4\) or \(x = 1.5\)
- C\(x = 4\) or \(x = -3\)
- D\(x = -4\) or \(x = 3\)
Solve \(x^{2} - 5x + 2 = 0\), giving your answers correct to 2 decimal places.
- A\(x = 4.56\) or \(x = 0.44\)
- B\(x = -4.56\) or \(x = -0.44\)
- C\(x = 5.37\) or \(x = -0.37\)
- D\(x = 7.06\) or \(x = 2.94\)
Solve \(x^{2} + 6x + 4 = 0\), giving your answers in the form \(p \pm \sqrt{q}\).
- A\(x = -3 \pm \sqrt{5}\)
- B\(x = 3 \pm \sqrt{5}\)
- C\(x = -3 \pm \sqrt{20}\)
- D\(x = -6 \pm \sqrt{20}\)
A right-angled triangle has shorter sides \(x\) cm and \((x + 7)\) cm, and hypotenuse 13 cm. Find the area of the triangle.
- A30 cm2
- B60 cm2
- C78 cm2
- D114 cm2
Quiz key
\(-6 \times 4 = -24\) and \(-6 + 4 = -2\), so \((x - 6)(x + 4) = 0\), giving \(x = 6\) or \(x = -4\). \(x = -6\) or \(x = 4\) reads the solutions straight off the brackets without changing sign. \(8\) and \(-3\) multiply to \(-24\) but add to 5.
\(3x^{2} - 12x = 0\), so \(3x(x - 4) = 0\), giving \(x = 0\) or \(x = 4\). Dividing by \(x\) loses the solution \(x = 0\), which is why '\(x = 4\) only' is wrong. \(x = \pm 2\) treats the equation as \(3x^{2} = 12\).
\(2x^{2} - 5x - 12 = (2x + 3)(x - 4)\). So \(2x + 3 = 0\) gives \(x = -1.5\) and \(x - 4 = 0\) gives \(x = 4\). \(x = -3\) forgets to divide by 2. \(x = -4\) or \(x = 1.5\) comes from \((2x - 3)(x + 4) = 2x^{2} + 5x - 12\), the wrong signs.
\(x = \dfrac{5 \pm \sqrt{25 - 8}}{2} = \dfrac{5 \pm \sqrt{17}}{2}\), so \(x = 4.56\) or \(0.44\) (2 d.p.). \(-4.56\) and \(-0.44\) use \(-b = -5\). \(5.37\) and \(-0.37\) use \(25 + 8\) under the root. \(7.06\) and \(2.94\) divide only the square root by 2, not the 5 as well.
\(x = \dfrac{-6 \pm \sqrt{36 - 16}}{2} = \dfrac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}\). \(-3 \pm \sqrt{20}\) halves the \(-6\) but not the root. \(-6 \pm \sqrt{20}\) does not divide by 2 at all. \(3 \pm \sqrt{5}\) uses \(-b = 6\).
\(x^{2} + (x + 7)^{2} = 169\) gives \(2x^{2} + 14x - 120 = 0\), so \(x^{2} + 7x - 60 = 0\), \((x + 12)(x - 5) = 0\). A length cannot be \(-12\), so \(x = 5\) and the sides are 5 cm and 12 cm. Area \(= \dfrac{1}{2} \times 5 \times 12 = 30\) cm2. 60 forgets the \(\dfrac{1}{2}\). 78 uses the hypotenuse as the height. 114 uses \(x = 12\), from the wrong sign in the bracket.
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.