The shape of the lesson
By the end of the lesson, learners can
- Substitute from the linear one: Elimination by adding or subtracting equations does not work when one equation has squared terms.
- Equations with an xy term: Some quadratic equations contain a product \(xy\).
- What the roots mean: Each root of the substituted quadratic is the \(x\)-coordinate (or \(y\)-coordinate) of a point where the line meets the curve.
- Rearrange the linear equation for the letter with the simplest coefficient.
- Substitute everywhere that letter appears in the quadratic, including \(xy\) terms.
- Expand squares fully, collect to \(ax^{2} + bx + c = 0\) and solve.
- Back-substitute each root into the LINEAR equation to find its partner.
- Pair the answers as coordinates; a repeated root means the line is a tangent.
One picture every idea returns to
Rearrange the LINEAR equation to make \(x\) or \(y\) the subject, substitute into the quadratic, solve the resulting quadratic, then substitute each root back into the linear equation. Solutions come in pairs \((x, y)\): one pair for each point of intersection.
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, Where a line meets a curve.. When one equation is linear and the other is quadratic, make one letter the subject of the linear equation and substitute. 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: Substitute from the linear one | Elimination by adding or subtracting equations does not work when one equation has squared terms. Model the worked example Line and parabola on the board, one line at a time. Your turn: Solve \(y = 3x - 2\) and \(y = x^{2}\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Equations with an xy term | Some quadratic equations contain a product \(xy\). Model the worked example Fractional solution on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: What the roots mean | Each root of the substituted quadratic is the \(x\)-coordinate (or \(y\)-coordinate) of a point where the line meets the curve. Model the worked example A tangent 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, 24 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 What the roots mean, 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 |
|---|---|
| \((-x - 1)^{2} = x^{2} + 2x + 1\), not \(x^{2} + 1\). | Write the square as a product and expand it fully before collecting terms. |
| Find each partner from the LINEAR equation, not the quadratic. | Substituting \(y = 2\) into \(x^{2} + 3y^{2} = 28\) gives \(x = \pm 4\), and \((-4, 2)\) is not a solution of \(x + 3y = 10\). |
| Give the answers as matched pairs. | Writing '\(x = 3\) or \(-2\), \(y = 5\) or \(0\)' does not say which \(y\) goes with which \(x\), and examiners withhold the final mark. |
Simultaneous equations: one linear, one quadratic: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
Solve the simultaneous equations \(y = 3x - 2\) and \(y = x^{2}\).
Solve the simultaneous equations \(x + 2y = 5\) and \(x^{2} + y^{2} = 10\).
Solve the simultaneous equations \(3x + y = 7\) and \(x^{2} + xy = 5\).
The line \(y = x - 1\) meets the curve \(y^{2} = x + 5\) at the points A and B.
(a) Find the coordinates of A and B.
(b) Find the coordinates of the midpoint of AB and the exact length of AB.
Show that the line \(y = 2x - 5\) and the curve \(y = x^{2} - 4x + 4\) meet at exactly one point, and find its coordinates.
A rectangular sheet of metal has perimeter 34 cm. The length of a diagonal of the sheet is 13 cm.
(a) Taking the sides as \(x\) cm and \(y\) cm, write down two equations in \(x\) and \(y\).
(b) Solve your equations to find the dimensions of the sheet, and hence its area.
Answer key and marking notes
- \(x^{2} = 3x - 2 \Rightarrow x^{2} - 3x + 2 = 0\).
- \((x - 1)(x - 2) = 0 \Rightarrow x = 1\) or \(x = 2\).
- \(x = 1,\ y = 1\) or \(x = 2,\ y = 4\).
- \(x = 5 - 2y\), so \((5 - 2y)^{2} + y^{2} = 10\).
- \(25 - 20y + 4y^{2} + y^{2} = 10 \Rightarrow 5y^{2} - 20y + 15 = 0 \Rightarrow y^{2} - 4y + 3 = 0\).
- \((y - 1)(y - 3) = 0 \Rightarrow y = 1\) or \(y = 3\).
- \(y = 1\): \(x = 3\). \(y = 3\): \(x = -1\). Solutions \((3, 1)\) and \((-1, 3)\).
- \(y = 7 - 3x\), so \(x^{2} + x(7 - 3x) = 5\).
- \(-2x^{2} + 7x - 5 = 0 \Rightarrow 2x^{2} - 7x + 5 = 0 \Rightarrow (2x - 5)(x - 1) = 0\).
- \(x = 1\): \(y = 4\).
- \(x = \dfrac{5}{2}\): \(y = 7 - \dfrac{15}{2} = -\dfrac{1}{2}\).
- (a) \((x - 1)^{2} = x + 5 \Rightarrow x^{2} - 2x + 1 = x + 5 \Rightarrow x^{2} - 3x - 4 = 0\).
- \((x - 4)(x + 1) = 0\): A\((4, 3)\) and B\((-1, -2)\).
- (b) Midpoint \(\left(\dfrac{4 + (-1)}{2},\ \dfrac{3 + (-2)}{2}\right) = \left(\dfrac{3}{2},\ \dfrac{1}{2}\right)\).
- AB \(= \sqrt{5^{2} + 5^{2}} = \sqrt{50} = 5\sqrt{2}\).
- \(x^{2} - 4x + 4 = 2x - 5 \Rightarrow x^{2} - 6x + 9 = 0\).
- Discriminant \(36 - 36 = 0\), so there is exactly one (repeated) root: the line is a tangent.
- \((x - 3)^{2} = 0 \Rightarrow x = 3,\ y = 2(3) - 5 = 1\). The point is \((3, 1)\).
- (a) \(2x + 2y = 34\), so \(x + y = 17\); by Pythagoras \(x^{2} + y^{2} = 169\).
- (b) \(y = 17 - x\): \(x^{2} + (17 - x)^{2} = 169 \Rightarrow 2x^{2} - 34x + 289 = 169\).
- \(2x^{2} - 34x + 120 = 0 \Rightarrow x^{2} - 17x + 60 = 0 \Rightarrow (x - 5)(x - 12) = 0\).
- \(x = 5,\ y = 12\) or \(x = 12,\ y = 5\): the sheet is 12 cm by 5 cm.
- Area \(= 12 \times 5 = 60\) cm2.
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 the simultaneous equations \(y = x + 3\) and \(y = x^{2} + 1\).
- A\((2,\ 5)\) and \((-1,\ 2)\)
- B\((-2,\ 1)\) and \((1,\ 4)\)
- C\((2,\ 5)\) only
- D\((2,\ 3)\) and \((-1,\ 0)\)
Substituting \(y = 2x - 1\) into \(x^{2} + y^{2} = 13\) gives which quadratic equation?
- A\(5x^{2} - 4x - 12 = 0\)
- B\(5x^{2} - 12 = 0\)
- C\(3x^{2} - 4x - 12 = 0\)
- D\(5x^{2} + 4x - 12 = 0\)
Solve the simultaneous equations \(x + y = 5\) and \(xy = 6\).
- A\((2,\ 3)\) and \((3,\ 2)\)
- B\((-2,\ -3)\) and \((-3,\ -2)\)
- C\((6,\ -1)\) and \((-1,\ 6)\)
- D\((1,\ 4)\) and \((4,\ 1)\)
A line and a curve are solved simultaneously. The quadratic equation obtained after substitution has a negative discriminant. What does this tell you?
- AThe line and the curve do not meet
- BThe line is a tangent to the curve
- CThe line crosses the curve at two points
- DThe line meets the curve at exactly one point and crosses it
Solve the simultaneous equations \(2x + y = 7\) and \(x^{2} + xy = 6\).
- A\((1,\ 5)\) and \((6,\ -5)\)
- B\((1,\ 5)\) and \((-6,\ 19)\)
- C\((-1,\ 9)\) and \((-6,\ 19)\)
- D\((1,\ 5)\) only
The line \(y = x - 1\) meets the circle \(x^{2} + y^{2} = 13\) at the points A and B. Find the length of AB.
- A\(5\sqrt{2}\)
- B\(\sqrt{26}\)
- C\(5\)
- D\(50\)
Which statement describes how the line \(y = 2x - 5\) meets the circle \(x^{2} + y^{2} = 5\)?
- AIt is a tangent to the circle at \((2,\ -1)\)
- BIt is a tangent to the circle at \((2,\ 1)\)
- CIt crosses the circle at two distinct points
- DIt does not meet the circle
A rectangle has perimeter 34 cm and its diagonals are 13 cm long. Find the area of the rectangle.
- A\(60\ \text{cm}^2\)
- B\(120\ \text{cm}^2\)
- C\(30\ \text{cm}^2\)
- D\(84.5\ \text{cm}^2\)
Quiz key
\(x^{2} + 1 = x + 3\) gives \(x^{2} - x - 2 = 0\), so \((x - 2)(x + 1) = 0\) and \(x = 2\) or \(x = -1\). From \(y = x + 3\): \((2,\ 5)\) and \((-1,\ 2)\). The tempting \((-2,\ 1)\), \((1,\ 4)\) solve \((x + 2)(x - 1) = 0\), a sign error in the factors. Keeping only \((2,\ 5)\) loses an intersection point, and \((2,\ 3)\), \((-1,\ 0)\) substitute into \(y = x + 1\), which is not one of the equations.
\((2x - 1)^{2} = 4x^{2} - 4x + 1\), so \(x^{2} + 4x^{2} - 4x + 1 = 13\), i.e. \(5x^{2} - 4x - 12 = 0\). (It factorises as \((x - 2)(5x + 6) = 0\), giving \((2,\ 3)\) and \(\left(-\frac{6}{5},\ -\frac{17}{5}\right)\).) The tempting \(5x^{2} - 12 = 0\) squares the bracket as \(4x^{2} + 1\), losing the middle term. \(3x^{2}\) uses \(2x^{2}\) for \((2x)^{2}\), and \(+4x\) has the wrong sign on the cross term.
\(x(5 - x) = 6\) gives \(x^{2} - 5x + 6 = 0\), so \(x = 2\) or \(x = 3\), and \(y = 5 - x\) gives the pairs \((2,\ 3)\) and \((3,\ 2)\). The tempting negative pair has product 6 but sum \(-5\). \((6,\ -1)\) has sum 5 but product \(-6\), from \(x^{2} - 5x - 6 = 0\), and \((1,\ 4)\) has product 4.
A negative discriminant means the quadratic has no real roots, so there is no point common to the line and the curve: they do not meet. The tempting "tangent" needs a ZERO discriminant (one repeated root). Two crossing points need a positive discriminant.
\(y = 7 - 2x\), so \(x^{2} + x(7 - 2x) = 6\), i.e. \(x^{2} + 7x - 2x^{2} = 6\), giving \(x^{2} - 7x + 6 = 0\) and \(x = 1\) or \(x = 6\). Then \(y = 7 - 2x\) gives \((1,\ 5)\) and \((6,\ -5)\). The tempting \((-6,\ 19)\) comes from expanding \(x(7 - 2x)\) as \(7x - 2x\), giving \(x^{2} + 5x - 6 = 0\). \((-1,\ 9)\), \((-6,\ 19)\) solve \(x^{2} + 7x + 6 = 0\), and \((1,\ 5)\) alone misses the second point.
\(x^{2} + (x - 1)^{2} = 13\) gives \(2x^{2} - 2x - 12 = 0\), so \(x^{2} - x - 6 = 0\) and \(x = 3\) or \(x = -2\). From \(y = x - 1\): A\((3,\ 2)\), B\((-2,\ -3)\). \(\text{AB} = \sqrt{5^{2} + 5^{2}} = \sqrt{50} = 5\sqrt{2}\). The tempting \(\sqrt{26}\) uses the wrong partner \(y = 3\) for \(x = -2\) (found from the circle, not the line). \(5\) uses only the change in \(x\), and \(50\) forgets the square root.
\(x^{2} + (2x - 5)^{2} = 5\) gives \(5x^{2} - 20x + 20 = 0\), i.e. \(x^{2} - 4x + 4 = (x - 2)^{2} = 0\): one repeated root, so the line is a tangent. From \(y = 2x - 5\), \(y = -1\), so the point of contact is \((2,\ -1)\). The tempting \((2,\ 1)\) takes \(y\) from \(y^{2} = 1\) in the circle equation, which allows \(\pm 1\); only \(-1\) lies on the line. A repeated root rules out two points or no points.
\(2x + 2y = 34\) gives \(x + y = 17\), and \(x^{2} + y^{2} = 169\). Substituting \(y = 17 - x\): \(2x^{2} - 34x + 120 = 0\), so \(x^{2} - 17x + 60 = 0\), giving sides 5 cm and 12 cm. Area \(= 5 \times 12 = 60\ \text{cm}^2\). The tempting \(120\) is \(2xy\), from \((x + y)^{2} - (x^{2} + y^{2}) = 289 - 169\), without halving. \(30\) is the area of the triangle cut off by a diagonal, and \(84.5\) assumes the rectangle is a square.
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.