The shape of the lesson
By the end of the lesson, learners can
- Three ways to solve: In the lab, pick a quadratic and solve it by factorising, then by the formula, and check that both routes land on the same roots.
- Inequalities: sketch first: A quadratic inequality is a question about the graph: where is the parabola above, or below, the \(x\)-axis?
- Surds, constants and sets of values: When the quadratic does not factorise, find the critical values with the formula and keep them exact.
- Rearrange so one side is zero; never divide by \(x\).
- Find the critical values by factorising, completing the square or the formula.
- Sketch the parabola through them (make \(a > 0\) by multiplying by \(-1\) and reversing the sign).
- Read: \(< 0\) between the roots, \(> 0\) outside, using \(\le\)/\(\ge\) to include the ends.
- Write an outside answer as two pieces joined by 'or'.
One picture every idea returns to
To solve a quadratic inequality, first solve the equation to find the critical values, then sketch the parabola. For \(a > 0\): \((x - \alpha)(x - \beta) < 0\) gives \(\alpha < x < \beta\) (between the roots); \((x - \alpha)(x - \beta) > 0\) gives \(x < \alpha\) or \(x > \beta\) (outside them).
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, Find the roots, then sketch.. Solve quadratic equations by factorising, completing the square or the formula, then use the roots and a quick sketch to solve quadratic inequalities, including the sets of values that discriminant questions produce. 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: Three ways to solve | In the lab, pick a quadratic and solve it by factorising, then by the formula, and check that both routes land on the same roots. Project tab 4 of the lab (Equations lab, quadratic equations) and let a student drive it while the class predicts. Model the worked example Factorising on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Inequalities: sketch first | A quadratic inequality is a question about the graph: where is the parabola above, or below, the \(x\)-axis? Model the worked example Between the roots on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Surds, constants and sets of values | When the quadratic does not factorise, find the critical values with the formula and keep them exact. Model the worked example Irrational critical values on the board, one line at a time. Your turn: Solve \(x(x + 3) \le 10\). | Cooperative Concept Mapping |
| 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, 20 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 Surds, constants and sets of values, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| Never divide both sides by \(x\): \(x^{2} = 4x\) becomes \(x = 4\) and the root \(x = 0\) is lost. | Rearrange to \(x^{2} - 4x = 0\), so \(x(x - 4) = 0\) and \(x = 0\) or \(x = 4\). |
| \(x^{2} > 9\) does NOT mean \(x > \pm 3\). | The curve \(y = x^{2} - 9\) is above the axis outside \(-3\) and \(3\), so \(x 3\). |
| An 'outside' answer is two separate pieces joined by 'or'. | Writing \(6 < k < -2\) is meaningless (no number is both bigger than 6 and smaller than \(-2\)) and loses the mark. |
Quadratic equations and inequalities: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Solve \(2x^{2} - 7x + 3 = 0\).
Solve \(5x^{2} - 3x - 1 = 0\), giving your answers correct to 3 significant figures.
Solve the inequality \(x^{2} - x - 20 \ge 0\).
Solve the inequality \((2x - 1)(x + 2) < 3x + 4\), giving your answer in exact form.
The equation \(kx^{2} + 2kx + 3 = 0\), where \(k\) is a non-zero constant, has no real roots. Find the set of possible values of \(k\).
A rectangle has length \((x + 5)\) cm and width \((x - 1)\) cm, where \(x > 1\). Its area is at least 40 cm2 and its perimeter is less than 32 cm.
Find the set of possible values of \(x\).
Answer key and marking notes
- \((2x - 1)(x - 3) = 0\).
- \(x = \dfrac{1}{2}\) or \(x = 3\).
- \(b^{2} - 4ac = 9 + 20 = 29\).
- \(x = \dfrac{3 \pm \sqrt{29}}{10}\).
- \(x = 0.839\) or \(x = -0.239\).
- \((x - 5)(x + 4) = 0\) gives critical values \(5\) and \(-4\).
- The parabola opens upwards, so \(\ge 0\) outside the roots, ends included.
- \(x \le -4\) or \(x \ge 5\).
- Expand: \(2x^{2} + 3x - 2 < 3x + 4\).
- Simplify: \(2x^{2} < 6 \Rightarrow x^{2} < 3\).
- \(-\sqrt{3} < x < \sqrt{3}\).
- No real roots: \((2k)^{2} - 4(k)(3) < 0\).
- \(4k^{2} - 12k < 0 \Rightarrow 4k(k - 3) < 0\).
- Critical values \(0\) and \(3\); a positive quadratic in \(k\) is negative between them.
- \(0 < k < 3\).
- Area: \((x + 5)(x - 1) \ge 40 \Rightarrow x^{2} + 4x - 45 \ge 0\).
- \((x + 9)(x - 5) \ge 0\), so \(x \le -9\) or \(x \ge 5\); with \(x > 1\) this gives \(x \ge 5\).
- Perimeter: \(2\big[(x + 5) + (x - 1)\big] = 4x + 8 < 32 \Rightarrow x < 6\).
- Both conditions: \(5 \le x < 6\).
8 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 \(3x^{2} - 5x - 2 = 0\).
- A\(x = 2\) or \(x = -\dfrac{1}{3}\)
- B\(x = -2\) or \(x = \dfrac{1}{3}\)
- C\(x = 2\) or \(x = \dfrac{1}{3}\)
- D\(x = 2\) or \(x = -3\)
Solve the inequality \(x^{2} - x - 12 \le 0\).
- A\(-3 \le x \le 4\)
- B\(x \le -3\) or \(x \ge 4\)
- C\(-4 \le x \le 3\)
- D\(-3 < x < 4\)
Solve the inequality \(x^{2} > 25\).
- A\(x < -5\) or \(x > 5\)
- B\(x > \pm 5\)
- C\(-5 < x < 5\)
- D\(x > 5\)
Solve \(2x^{2} - 4x - 3 = 0\), giving your answers in exact form.
- A\(x = 1 \pm \dfrac{\sqrt{10}}{2}\)
- B\(x = 2 \pm \sqrt{10}\)
- C\(x = -1 \pm \dfrac{\sqrt{10}}{2}\)
- DThere are no real roots
Solve the inequality \(3x^{2} + 7x \ge 6\).
- A\(x \le -3\) or \(x \ge \dfrac{2}{3}\)
- B\(-3 \le x \le \dfrac{2}{3}\)
- C\(x \le -\dfrac{2}{3}\) or \(x \ge 3\)
- D\(x < -3\) or \(x > \dfrac{2}{3}\)
Solve the inequality \(6 + x - 2x^{2} < 0\).
- A\(x < -\dfrac{3}{2}\) or \(x > 2\)
- B\(-\dfrac{3}{2} < x < 2\)
- C\(x < -2\) or \(x > \dfrac{3}{2}\)
- D\(-2 < x < \dfrac{3}{2}\)
Solve the inequality \(x^{2} - 6x + 4 > 0\), giving your answer in exact form.
- A\(x < 3 - \sqrt{5}\) or \(x > 3 + \sqrt{5}\)
- B\(3 - \sqrt{5} < x < 3 + \sqrt{5}\)
- C\(x < -3 - \sqrt{5}\) or \(x > -3 + \sqrt{5}\)
- D\(x < 3 - \sqrt{13}\) or \(x > 3 + \sqrt{13}\)
Find the set of values of \(k\) for which the equation \(x^{2} + (k - 1)x + 9 = 0\) has no real roots.
- A\(-5 < k < 7\)
- B\(k < -5\) or \(k > 7\)
- C\(-7 < k < 5\)
- D\(-35 < k < 37\)
Quiz key
\(3x^{2} - 5x - 2 = (3x + 1)(x - 2)\), so \(3x + 1 = 0\) or \(x - 2 = 0\), giving \(x = -\dfrac{1}{3}\) or \(x = 2\). The tempting \(x = -2\) or \(\dfrac{1}{3}\) reads the numbers in the brackets with the wrong signs. \(x = -3\) comes from setting \(x + 3 = 0\) instead of solving \(3x + 1 = 0\).
\(x^{2} - x - 12 = (x - 4)(x + 3)\), with critical values \(-3\) and \(4\). The parabola opens upwards, so it is on or below the axis BETWEEN the roots: \(-3 \le x \le 4\). The tempting outside region is where the curve is above the axis (\(\ge 0\)). \(-4 \le x \le 3\) has the signs of the roots swapped, and strict inequalities lose the end points, where the expression equals 0.
\(x^{2} - 25 = (x - 5)(x + 5) > 0\). The parabola is above the axis OUTSIDE its roots, so \(x < -5\) or \(x > 5\). The tempting \(x > \pm 5\) is meaningless: \(x = -6\) satisfies \(x^{2} > 25\) but is not greater than \(-5\). \(-5 < x < 5\) solves \(x^{2} < 25\), and \(x > 5\) misses the negative values.
\(x = \dfrac{4 \pm \sqrt{16 + 24}}{4} = \dfrac{4 \pm \sqrt{40}}{4} = \dfrac{4 \pm 2\sqrt{10}}{4} = 1 \pm \dfrac{\sqrt{10}}{2}\). The tempting \(2 \pm \sqrt{10}\) divides by 2 instead of \(2a = 4\). \(-1 \pm\) uses \(-b = -4\), and "no real roots" comes from \(16 - 24\), taking \(c = +3\).
\(3x^{2} + 7x - 6 = (3x - 2)(x + 3) \ge 0\), with critical values \(-3\) and \(\dfrac{2}{3}\). The parabola opens upwards, so it is on or above the axis outside the roots: \(x \le -3\) or \(x \ge \dfrac{2}{3}\). The tempting between-region solves \(\le 0\). \(-\dfrac{2}{3}\) and \(3\) have the signs of the roots swapped, and strict inequalities wrongly exclude the critical values, where the two sides are equal.
Multiplying by \(-1\): \(2x^{2} - x - 6 > 0\), so \((2x + 3)(x - 2) > 0\). Critical values \(-\dfrac{3}{2}\) and \(2\); an upward parabola is positive outside them: \(x < -\dfrac{3}{2}\) or \(x > 2\). (Equivalently, the downward curve \(y = 6 + x - 2x^{2}\) is below the axis outside its roots.) The tempting \(-\dfrac{3}{2} < x < 2\) treats the original curve as opening upwards. The other two options use the roots of \((2x - 3)(x + 2)\), a factorising sign error.
\(x^{2} - 6x + 4 = (x - 3)^{2} - 5\), so the critical values are \(3 \pm \sqrt{5}\). The parabola opens upwards and is positive outside them: \(x < 3 - \sqrt{5}\) or \(x > 3 + \sqrt{5}\). The tempting between-region solves \(< 0\). \(-3 \pm \sqrt{5}\) uses the wrong sign in the bracket, and \(3 \pm \sqrt{13}\) comes from \(\sqrt{36 + 16}\), taking \(c = -4\).
\((k - 1)^{2} - 36 < 0\), so \((k - 1)^{2} < 36\), i.e. \(-6 < k - 1 < 6\), giving \(-5 < k < 7\). (Or \((k - 7)(k + 5) < 0\): between the roots.) The tempting outside region gives two distinct real roots. \(-7 < k < 5\) subtracts 1 instead of adding it, and \(-35 < k < 37\) forgets to take the square root of 36.
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.