The shape of the lesson
By the end of the lesson, learners can
- Spot the hidden quadratic: Look for two terms where one power is exactly double the other, plus a constant.
- Roots and reciprocals: With \(\sqrt{x}\), remember that the symbol means the positive square root, so a negative value of \(u = \sqrt{x}\) must be rejected.
- Other functions of x: The same idea works for any function: fractional powers, powers of a number and trigonometric functions.
- Spot two terms where one power is double the other.
- Substitute \(u\) for the smaller power (or the function) and solve the quadratic in \(u\).
- Reject values the function can never take: negative \(x^{2}\), negative \(\sqrt{x}\), \(\sin x\) outside \([-1, 1]\).
- Undo the substitution for each remaining value, including both signs for \(x^{2} = k\).
- Check in the original equation when a square root is involved.
One picture every idea returns to
If an equation has the shape \(a\big[f(x)\big]^{2} + b\,f(x) + c = 0\), put \(u = f(x)\), solve \(au^{2} + bu + c = 0\), then solve \(f(x) = u\) for each root. Reject any value that \(f(x)\) can never take, such as a negative \(x^{2}\) or a negative \(\sqrt{x}\).
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, A quadratic in disguise.. \(x^{4} - 5x^{2} + 4 = 0\) looks like a quartic, but it is really a quadratic in \(x^{2}\). 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: Spot the hidden quadratic | Look for two terms where one power is exactly double the other, plus a constant. Model the worked example Four roots on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Roots and reciprocals | With \(\sqrt{x}\), remember that the symbol means the positive square root, so a negative value of \(u = \sqrt{x}\) must be rejected. Model the worked example Square root on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Other functions of x | The same idea works for any function: fractional powers, powers of a number and trigonometric functions. Model the worked example Cube roots on the board, one line at a time. Your turn: Solve \(2^{2x} - 6 \times 2^{x} + 8 = 0\). | 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, 21 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 Other functions of x, 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 |
|---|---|
| Stopping at \(u = 1\) or \(u = 4\) answers the wrong question. | The equation was in \(x\), so convert back, and remember that \(x^{2} = 4\) has TWO solutions, \(\pm 2\). |
| Check answers in the ORIGINAL equation. | For \(6x - \sqrt{x} - 2 = 0\), \(x = \frac{1}{4}\) (from \(u = -\frac{1}{2}\)) gives \(1.5 - 0.5 - 2 = -1 \ne 0\), which confirms it must be rejected. |
| Do not reject a negative value of a cube root: \(x^{\frac{1}{3}} = -4\) is fine and gives \(x = -64\). | Only reject values the function can never take (negative \(\sqrt{x}\) or \(x^{2}\), \(\sin x\) outside \(-1\) to \(1\), negative \(2^{x}\)). |
Equations quadratic in a function of x: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Solve \(x^{4} - 10x^{2} + 9 = 0\).
Solve \(x - 6\sqrt{x} + 8 = 0\).
Find the real values of \(x\) for which \(4x^{4} + 7x^{2} - 2 = 0\).
Solve \(\dfrac{6}{x^{2}} - \dfrac{1}{x} - 1 = 0\).
Find the coordinates of the points where the curves \(y = x^{2} + 2\) and \(y = \dfrac{3}{x^{2}}\) intersect.
A right-angled triangle has area 2 cm2. One of the shorter sides is \(x\) cm and the hypotenuse is \(\sqrt{17}\) cm.
(a) Show that the other shorter side is \(\dfrac{4}{x}\) cm and that \(x^{4} - 17x^{2} + 16 = 0\).
(b) Find the lengths of the two shorter sides.
Answer key and marking notes
- Let \(u = x^{2}\): \(u^{2} - 10u + 9 = 0 \Rightarrow (u - 1)(u - 9) = 0\).
- \(x^{2} = 1\) or \(x^{2} = 9\).
- \(x = \pm 1\) or \(x = \pm 3\).
- Let \(u = \sqrt{x}\): \(u^{2} - 6u + 8 = 0 \Rightarrow (u - 2)(u - 4) = 0\).
- \(\sqrt{x} = 2\) or \(\sqrt{x} = 4\), both positive, so both valid.
- \(x = 4\) or \(x = 16\).
- Let \(u = x^{2}\): \(4u^{2} + 7u - 2 = 0 \Rightarrow (4u - 1)(u + 2) = 0\).
- \(x^{2} = -2\) has no real solutions, so reject it.
- \(x^{2} = \dfrac{1}{4} \Rightarrow x = \pm\dfrac{1}{2}\).
- Let \(u = \dfrac{1}{x}\): \(6u^{2} - u - 1 = 0 \Rightarrow (3u + 1)(2u - 1) = 0\).
- \(u = \dfrac{1}{2} \Rightarrow x = 2\).
- \(u = -\dfrac{1}{3} \Rightarrow x = -3\).
- \(x^{2} + 2 = \dfrac{3}{x^{2}} \Rightarrow x^{4} + 2x^{2} - 3 = 0\).
- Let \(u = x^{2}\): \((u + 3)(u - 1) = 0\); \(u = -3\) is impossible.
- \(x^{2} = 1 \Rightarrow x = \pm 1\), and \(y = 1 + 2 = 3\).
- The curves meet at \((1, 3)\) and \((-1, 3)\).
- (a) Area \(= \dfrac{1}{2} \times x \times y = 2 \Rightarrow y = \dfrac{4}{x}\).
- Pythagoras: \(x^{2} + \dfrac{16}{x^{2}} = 17\); multiply by \(x^{2}\): \(x^{4} - 17x^{2} + 16 = 0\).
- (b) Let \(u = x^{2}\): \((u - 1)(u - 16) = 0\), so \(x^{2} = 1\) or \(16\).
- \(x\) is a length, so \(x > 0\): \(x = 1\) or \(x = 4\).
- Either way the shorter sides are 1 cm and 4 cm (check: \(1 + 16 = 17\), area \(\frac{1}{2} \times 1 \times 4 = 2\)).
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 \(x^{4} - 10x^{2} + 9 = 0\).
- A\(x = \pm 1\) or \(x = \pm 3\)
- B\(x = 1\) or \(x = 9\)
- C\(x = 1\) or \(x = 3\)
- D\(x = \pm 1\) only
Solve \(x^{6} - 9x^{3} + 8 = 0\) for real \(x\).
- A\(x = 1\) or \(x = 2\)
- B\(x = 1\) or \(x = 8\)
- C\(x = \pm 1\) or \(x = \pm 2\)
- D\(x = 1\) or \(x = 512\)
Solve \(x - 3\sqrt{x} - 4 = 0\).
- A\(x = 16\)
- B\(x = 16\) or \(x = 1\)
- C\(x = 4\)
- D\(x = 4\) or \(x = -1\)
Solve \(\dfrac{2}{x^{2}} + \dfrac{1}{x} - 6 = 0\).
- A\(x = \dfrac{2}{3}\) or \(x = -\dfrac{1}{2}\)
- B\(x = \dfrac{3}{2}\) or \(x = -2\)
- C\(x = -\dfrac{2}{3}\) or \(x = \dfrac{1}{2}\)
- D\(x = \dfrac{2}{3}\) only
Find the exact real solutions of \(x^{4} + 2x^{2} - 15 = 0\).
- A\(x = \pm\sqrt{3}\)
- B\(x = \pm\sqrt{3}\) or \(x = \pm\sqrt{5}\)
- C\(x = 3\)
- D\(x = \pm\sqrt{5}\)
Solve \(x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 4 = 0\).
- A\(x = 1\) or \(x = -64\)
- B\(x = 1\) only
- C\(x = 1\) or \(x = 64\)
- D\(x = 1\) or \(x = 16\)
Solve \(3^{2x} - 10 \times 3^{x} + 9 = 0\).
- A\(x = 0\) or \(x = 2\)
- B\(x = 1\) or \(x = 9\)
- C\(x = 1\) or \(x = 2\)
- D\(x = 2\) only
Solve \(2\cos^{2} x - 3\cos x + 1 = 0\) for \(0^{\circ} \le x \le 360^{\circ}\).
- A\(x = 0^{\circ},\ 60^{\circ},\ 300^{\circ},\ 360^{\circ}\)
- B\(x = 60^{\circ},\ 300^{\circ}\)
- C\(x = 0^{\circ},\ 60^{\circ},\ 360^{\circ}\)
- D\(x = 0^{\circ},\ 60^{\circ},\ 120^{\circ},\ 360^{\circ}\)
Quiz key
With \(u = x^{2}\): \(u^{2} - 10u + 9 = 0\), so \((u - 1)(u - 9) = 0\) and \(u = 1\) or \(u = 9\). Then \(x^{2} = 1\) gives \(x = \pm 1\) and \(x^{2} = 9\) gives \(x = \pm 3\): four roots. The tempting \(x = 1\) or \(9\) stops at the values of \(u\). \(x = 1\) or \(3\) forgets the negative square roots, and \(\pm 1\) only loses the second factor.
Let \(u = x^{3}\): \(u^{2} - 9u + 8 = 0\), so \(u = 1\) or \(u = 8\). Then \(x^{3} = 1\) gives \(x = 1\) and \(x^{3} = 8\) gives \(x = 2\). A cube has only one real cube root, so there are no \(\pm\) pairs. The tempting \(x = 1\) or \(8\) stops at \(u\). \(\pm\) wrongly treats a cube root like a square root (\((-2)^{3} = -8\), not 8), and 512 cubes \(u\) instead of taking the cube root.
With \(u = \sqrt{x}\): \(u^{2} - 3u - 4 = 0\), so \((u - 4)(u + 1) = 0\) and \(\sqrt{x} = 4\) or \(\sqrt{x} = -1\). The positive square root can never be negative, so reject \(-1\). \(\sqrt{x} = 4\) gives \(x = 16\). Check: \(16 - 12 - 4 = 0\). The tempting \(x = 1\) squares the rejected value: \(1 - 3 - 4 = -6 \ne 0\). \(x = 4\) and \(x = -1\) stop at the values of \(\sqrt{x}\).
With \(u = \dfrac{1}{x}\): \(2u^{2} + u - 6 = 0\), so \((2u - 3)(u + 2) = 0\) and \(u = \dfrac{3}{2}\) or \(u = -2\). Taking reciprocals: \(x = \dfrac{2}{3}\) or \(x = -\dfrac{1}{2}\). Both are valid, since \(\dfrac{1}{x}\) can be negative. The tempting \(\dfrac{3}{2}\) and \(-2\) are the values of \(u\), not \(x\). The third option has sign errors, and rejecting the negative root is wrong here.
\(u^{2} + 2u - 15 = 0\) gives \((u + 5)(u - 3) = 0\), so \(x^{2} = 3\) or \(x^{2} = -5\). A real square is never negative, so reject \(-5\). \(x^{2} = 3\) gives \(x = \pm\sqrt{3}\). The tempting \(\pm\sqrt{5}\) comes from writing \(x^{2} = 5\), dropping the minus sign on the rejected root (check: \(25 + 10 - 15 = 20 \ne 0\)). \(x = 3\) stops at \(u\).
\(u^{2} + 3u - 4 = 0\) gives \((u + 4)(u - 1) = 0\), so \(x^{\frac{1}{3}} = 1\) or \(x^{\frac{1}{3}} = -4\). Cubing: \(x = 1\) or \(x = (-4)^{3} = -64\). Cube roots of negative numbers exist, so both are valid. Check \(x = -64\): \(16 - 12 - 4 = 0\). The tempting \(x = 1\) only wrongly rejects the negative value, as if it were a square root. 64 loses the sign of \((-4)^{3}\), and 16 squares instead of cubing.
With \(u = 3^{x}\): \(u^{2} - 10u + 9 = 0\), so \(u = 1\) or \(u = 9\). \(3^{x} = 1\) gives \(x = 0\), and \(3^{x} = 9\) gives \(x = 2\). The tempting \(x = 1\) or \(9\) stops at \(u\). \(x = 1\) from \(3^{x} = 1\) forgets that any nonzero number to the power 0 is 1, and \(x = 2\) only discards a valid root.
\(2u^{2} - 3u + 1 = (2u - 1)(u - 1) = 0\), so \(\cos x = \dfrac{1}{2}\) or \(\cos x = 1\). \(\cos x = \dfrac{1}{2}\): \(x = 60^{\circ}\) or \(360^{\circ} - 60^{\circ} = 300^{\circ}\). \(\cos x = 1\): \(x = 0^{\circ}\) or \(360^{\circ}\), both inside the closed interval. The tempting \(60^{\circ},\ 300^{\circ}\) ignores the root \(\cos x = 1\). Leaving out \(300^{\circ}\) misses the second angle with cosine \(\dfrac{1}{2}\), and \(120^{\circ}\) has cosine \(-\dfrac{1}{2}\).
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.