The shape of the lesson
By the end of the lesson, learners can
- Simplifying surds: In the lab, choose Simplify and move the slider through numbers such as 20, 72 and 200.
- Adding, subtracting and expanding: Like surds collect like algebra: \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\), just as \(3x + 5x = 8x\).
- Rationalising the denominator: An answer with a surd on the bottom is not in its simplest form.
- Simplify: split off the largest square factor: \(\sqrt{98} = 7\sqrt{2}\).
- Collect only like surds, after simplifying each one.
- Expand every term; \(\sqrt{a} \times \sqrt{a} = a\).
- Rationalise \(\dfrac{k}{\sqrt{a}}\) by multiplying by \(\dfrac{\sqrt{a}}{\sqrt{a}}\).
- Conjugate: for \(p + \sqrt{q}\) on the bottom, multiply by \(p - \sqrt{q}\).
One picture every idea returns to
Use \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\) with \(a\) the largest square factor: \(\sqrt{20} = \sqrt{4}\sqrt{5} = 2\sqrt{5}\). To rationalise, multiply top and bottom by \(\sqrt{a}\), or by the conjugate: \((\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2\).
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, Keep it exact.. 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: Simplifying surds | In the lab, choose Simplify and move the slider through numbers such as 20, 72 and 200. Project tab 6 of the lab (Powers and accuracy lab, surds) and let a student drive it while the class predicts. Model the worked example Take out the largest square on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Adding, subtracting and expanding | Like surds collect like algebra: \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\), just as \(3x + 5x = 8x\). Model the worked example Simplify, then collect on the board, one line at a time. Your turn: Simplify \(\sqrt{12} + \sqrt{27} - \sqrt{48}\). | Pairs |
| 21 to 30 min | Teach and check: Rationalising the denominator | An answer with a surd on the bottom is not in its simplest form. Model the worked example A single surd on the bottom on the board, one line at a time. | Round Table |
| 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, 15 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 Rationalising the denominator, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| \(\sqrt{72} = 2\sqrt{18}\) is true but NOT fully simplified, because 18 still has the square factor 9. | Use the LARGEST square factor, or keep going until no square factor is left. |
| \(\sqrt{200} - \sqrt{32}\) is NOT \(\sqrt{168}\). | You can only subtract surds once they are the same surd, like \(10\sqrt{2} - 4\sqrt{2}\). |
| Multiplying \(\dfrac{1}{\sqrt{3} - 1}\) by \(\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\) does not help: the bottom becomes \(4 - 2\sqrt{3}\), still a surd. | Use the conjugate, with the opposite sign. |
Surds: simplifying and rationalising: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Simplify (a) \(\sqrt{45}\) (b) \(\sqrt{98}\).
Simplify \(\sqrt{12} + \sqrt{27} - \sqrt{48}\).
Expand and simplify \((5 - \sqrt{2})(3 + \sqrt{2})\).
Rationalise the denominator and simplify (a) \(\dfrac{15}{\sqrt{5}}\) (b) \(\dfrac{2}{\sqrt{8}}\).
Rationalise the denominator of \(\dfrac{6}{\sqrt{5} - \sqrt{2}}\). Give your answer in its simplest form.
A right-angled triangle has shorter sides \(\sqrt{12}\) cm and \(\sqrt{24}\) cm.
(a) Find the length of the hypotenuse.
(b) Find the area in the form \(a\sqrt{2}\) cm2.
(c) Find the perimeter, with each surd in its simplest form.
Answer key and marking notes
- (a) \(\sqrt{9}\sqrt{5} = 3\sqrt{5}\)
- (b) \(\sqrt{49}\sqrt{2} = 7\sqrt{2}\)
- \(2\sqrt{3} + 3\sqrt{3} - 4\sqrt{3}\)
- \(= \sqrt{3}\)
- \(15 + 5\sqrt{2} - 3\sqrt{2} - 2\)
- \(= 13 + 2\sqrt{2}\)
- (a) \(\dfrac{15\sqrt{5}}{5} = 3\sqrt{5}\)
- (b) \(\sqrt{8} = 2\sqrt{2}\), so \(\dfrac{2}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
- Multiply top and bottom by \(\sqrt{5} + \sqrt{2}\).
- Bottom: \((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) = 5 - 2 = 3\).
- \(\dfrac{6(\sqrt{5} + \sqrt{2})}{3} = 2\sqrt{5} + 2\sqrt{2}\)
- (a) \(h^{2} = 12 + 24 = 36\), so \(h = 6\) cm.
- (b) Area \(= \dfrac{1}{2}\sqrt{12}\sqrt{24} = \dfrac{1}{2}\sqrt{288} = \dfrac{1}{2} \times 12\sqrt{2} = 6\sqrt{2}\) cm2.
- (c) \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{24} = 2\sqrt{6}\).
- Perimeter \(= 6 + 2\sqrt{3} + 2\sqrt{6}\) cm.
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.
Simplify \(\sqrt{98}\).
- A\(7\sqrt{2}\)
- B\(49\sqrt{2}\)
- C\(2\sqrt{7}\)
- D\(14\sqrt{2}\)
Simplify \(\sqrt{108} + \sqrt{48}\).
- A\(10\sqrt{3}\)
- B\(2\sqrt{39}\)
- C\(10\sqrt{6}\)
- D\(24\sqrt{3}\)
Expand and simplify \((3 - \sqrt{5})^{2}\).
- A\(14 - 6\sqrt{5}\)
- B\(4\)
- C\(4 - 6\sqrt{5}\)
- D\(14 - 3\sqrt{5}\)
Expand and simplify \((4 + \sqrt{7})(4 - \sqrt{7})\).
- A\(9\)
- B\(23\)
- C\(9 - 8\sqrt{7}\)
- D\(16 - \sqrt{7}\)
Rationalise the denominator of \(\dfrac{12}{\sqrt{6}}\), giving your answer in its simplest form.
- A\(2\sqrt{6}\)
- B\(12\sqrt{6}\)
- C\(\dfrac{\sqrt{6}}{2}\)
- D\(2\)
Write \(\dfrac{8}{3 - \sqrt{5}}\) in the form \(a + b\sqrt{5}\).
- A\(6 + 2\sqrt{5}\)
- B\(6 - 2\sqrt{5}\)
- C\(24 + 8\sqrt{5}\)
- D\(\dfrac{12 + 4\sqrt{5}}{7}\)
Quiz key
\(98 = 49 \times 2\), so \(\sqrt{98} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}\). The tempting \(49\sqrt{2}\) takes the square factor outside without square-rooting it. \(2\sqrt{7}\) swaps the roles of the numbers, and \(14\sqrt{2}\) doubles 7.
\(\sqrt{108} = \sqrt{36}\sqrt{3} = 6\sqrt{3}\) and \(\sqrt{48} = \sqrt{16}\sqrt{3} = 4\sqrt{3}\), so the sum is \(10\sqrt{3}\). The tempting \(2\sqrt{39}\) adds under the root (\(\sqrt{156}\)), but \(\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}\). \(10\sqrt{6}\) adds the roots too, and \(24\sqrt{3}\) multiplies 6 and 4.
\((3 - \sqrt{5})(3 - \sqrt{5}) = 9 - 3\sqrt{5} - 3\sqrt{5} + 5 = 14 - 6\sqrt{5}\). The tempting 4 squares each term separately (\(9 - 5\)) and misses the middle terms. \(4 - 6\sqrt{5}\) treats \((\sqrt{5})^{2}\) as \(-5\), and \(14 - 3\sqrt{5}\) counts only one middle term.
\((4 + \sqrt{7})(4 - \sqrt{7}) = 16 - 4\sqrt{7} + 4\sqrt{7} - 7 = 9\). The surd terms cancel. The tempting 23 adds the 7 instead of subtracting it (\(+\sqrt{7} \times -\sqrt{7} = -7\)). \(9 - 8\sqrt{7}\) makes both middle terms negative, and \(16 - \sqrt{7}\) multiplies only the first and last terms.
\(\dfrac{12}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} = \dfrac{12\sqrt{6}}{6} = 2\sqrt{6}\). The tempting \(12\sqrt{6}\) multiplies the top by \(\sqrt{6}\) but forgets the bottom becomes 6. \(\dfrac{\sqrt{6}}{2}\) is the reciprocal of the answer, and 2 cancels the surd as if \(\dfrac{\sqrt{6}}{6} = \dfrac{1}{6}\).
Multiply by \(\dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}\): the bottom is \(9 - 5 = 4\), the top is \(24 + 8\sqrt{5}\), so the answer is \(\dfrac{24 + 8\sqrt{5}}{4} = 6 + 2\sqrt{5}\). The tempting \(6 - 2\sqrt{5}\) keeps the minus sign from the denominator in the conjugate's place. \(24 + 8\sqrt{5}\) forgets to divide by 4, and \(\dfrac{12 + 4\sqrt{5}}{7}\) uses \(9 + 5 = 14\) for the bottom.
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.