The shape of the lesson
By the end of the lesson, learners can
- One number, three outcomes: The quadratic formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) only gives real answers when the number under the root is not negative.
- Finding an unknown constant: Most exam questions give a quadratic containing a constant \(k\) and describe its roots.
- Always positive, always real: The discriminant also proves facts that hold for every \(x\) or every value of a constant.
- Rearrange to \(ax^{2} + bx + c = 0\) and read \(a\), \(b\), \(c\) with their signs.
- Translate the words: equal \(= 0\); distinct \(> 0\); real \(\ge 0\); no real \(< 0\).
- Substitute with brackets: \((b)^{2} - 4(a)(c)\).
- Solve the resulting equation or inequality in \(k\); a quadratic inequality has a two-ended answer.
- Always positive needs \(a > 0\) as well as \(b^{2} - 4ac < 0\).
One picture every idea returns to
For \(ax^{2} + bx + c = 0\): if \(b^{2} - 4ac > 0\) there are two distinct real roots; if \(b^{2} - 4ac = 0\) there is one repeated root; if \(b^{2} - 4ac < 0\) there are no real roots. 'Real roots' (not necessarily distinct) means \(b^{2} - 4ac \ge 0\).
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, Count the roots before you solve.. Hidden under the square root of the quadratic formula is one number, \(b^{2} - 4ac\). 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: One number, three outcomes | The quadratic formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) only gives real answers when the number under the root is not negative. Model the worked example Classify three equations on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Finding an unknown constant | Most exam questions give a quadratic containing a constant \(k\) and describe its roots. Model the worked example Equal roots on the board, one line at a time. Your turn: Find the set of values of \(k\) for which \(2x^{2} - 8x + k = 0\) has two distinct real roots. | Pairs |
| 21 to 30 min | Teach and check: Always positive, always real | The discriminant also proves facts that hold for every \(x\) or every value of a constant. Model the worked example Positive for all x 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, 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 Always positive, always real, 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 |
|---|---|
| \(b^{2}\) is never negative: with \(b = -5\), \(b^{2} = 25\), not \(-25\). | Bracket negative values when you substitute: \((-5)^{2} - 4(3)(1)\). |
| Read the wording exactly. 'Real roots' includes equal roots, so it needs \(\ge 0\); 'two distinct real roots' needs strict \(> 0\). | Using the wrong one costs the final mark. |
| For 'positive for all \(x\)' the condition \(b^{2} - 4ac 0\). | Return to Always positive, always real: ask a pair to state the correct version and show it on the lab before moving on. |
The discriminant: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Find the discriminant of \(5x^{2} - 2x + 3\) and hence state the number of real roots of \(5x^{2} - 2x + 3 = 0\).
Find the values of \(k\) for which \(4x^{2} + kx + 9 = 0\) has a repeated root. For the positive value of \(k\), find the root.
Find the set of values of \(k\) for which the equation \(x^{2} + 6x + 2k - 1 = 0\) has two distinct real roots.
Find the set of values of \(k\) for which \(kx^{2} - 4x + k > 0\) for all real values of \(x\).
Show that the equation \(2x^{2} + (p - 4)x - p = 0\) has two distinct real roots for all real values of \(p\).
A rectangular garden is to be fenced with exactly 40 m of fencing. One side has length \(x\) m and the area is \(A\) m2.
(a) Show that \(x^{2} - 20x + A = 0\).
(b) Use the discriminant to find the greatest possible area, and the dimensions of the garden in that case.
(c) Explain why an area of 105 m2 is impossible.
Answer key and marking notes
- \((-2)^{2} - 4(5)(3) = 4 - 60 = -56\).
- \(-56 < 0\), so there are no real roots.
- \(k^{2} - 4(4)(9) = 0 \Rightarrow k^{2} = 144 \Rightarrow k = \pm 12\).
- \(k = 12\): \(4x^{2} + 12x + 9 = (2x + 3)^{2} = 0\).
- The repeated root is \(x = -\dfrac{3}{2}\).
- \(36 - 4(1)(2k - 1) > 0\).
- \(36 - 8k + 4 > 0 \Rightarrow 40 > 8k\).
- \(k < 5\).
- The parabola must open upwards: \(k > 0\).
- It must not meet the \(x\)-axis: \(16 - 4k^{2} < 0 \Rightarrow k^{2} > 4\).
- \(k^{2} > 4 \Rightarrow k < -2\) or \(k > 2\).
- Combined with \(k > 0\): \(k > 2\).
- \(b^{2} - 4ac = (p - 4)^{2} - 4(2)(-p) = p^{2} - 8p + 16 + 8p\).
- \(= p^{2} + 16\).
- \(p^{2} \ge 0\), so \(p^{2} + 16 \ge 16 > 0\): two distinct real roots for every \(p\).
- (a) The other side is \(\dfrac{40 - 2x}{2} = 20 - x\), so \(A = x(20 - x) = 20x - x^{2}\).
- Rearranging: \(x^{2} - 20x + A = 0\).
- (b) For a real length \(x\), \(b^{2} - 4ac \ge 0\): \(400 - 4A \ge 0 \Rightarrow A \le 100\).
- Greatest area 100 m2; then \(x^{2} - 20x + 100 = (x - 10)^{2} = 0\), so the garden is a 10 m by 10 m square.
- (c) \(A = 105\) gives \(400 - 420 = -20 < 0\): no real value of \(x\), so this area cannot be made.
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.
Find the value of the discriminant of \(3 - 4x - 2x^{2}\).
- A\(40\)
- B\(-8\)
- C\(8\)
- D\(-40\)
Which statement about the roots of \(9x^{2} - 30x + 25 = 0\) is correct?
- AOne repeated root, \(x = \dfrac{5}{3}\)
- BTwo distinct real roots
- CNo real roots
- DOne repeated root, \(x = -\dfrac{5}{3}\)
The equation \(x^{2} - kx + 9 = 0\) has equal roots. Find the possible values of \(k\).
- A\(k = 6\) or \(k = -6\)
- B\(k = 6\) only
- C\(k = 3\) or \(k = -3\)
- D\(k = 36\)
Find the set of values of \(k\) for which \(3x^{2} + 6x + k = 0\) has no real roots.
- A\(k > 3\)
- B\(k < 3\)
- C\(k \ge 3\)
- D\(k > 12\)
Find the set of values of \(k\) for which the equation \(x^{2} + kx + 2k - 3 = 0\) has two distinct real roots.
- A\(k < 2\) or \(k > 6\)
- B\(2 < k < 6\)
- C\(k \le 2\) or \(k \ge 6\)
- D\(k < -6\) or \(k > -2\)
The line \(y = mx - 2\) is a tangent to the curve \(y = x^{2} + 2x + 2\). Find the possible values of the constant \(m\).
- A\(m = 6\) or \(m = -2\)
- B\(m = -6\) or \(m = 2\)
- C\(m = 6\) only
- D\(m = 2 \pm 2\sqrt{2}\)
Find the set of values of \(k\) for which \(kx^{2} - 4x + k < 0\) for all real values of \(x\).
- A\(k < -2\)
- B\(k > 2\)
- C\(k < -2\) or \(k > 2\)
- D\(-2 < k < 0\)
Which statement about the roots of \(x^{2} + (p + 2)x + p = 0\) is true for all real values of the constant \(p\)?
- AThe discriminant is \(p^{2} + 4\), so there are always two distinct real roots
- BThe discriminant is \((p - 2)^{2}\), so there are always real roots, equal when \(p = 2\)
- CThe discriminant is \(p^{2} + 4\), so the roots are equal when \(p = 0\)
- DThe discriminant depends on \(p\), so for some values of \(p\) there are no real roots
Quiz key
\(-2x^{2} - 4x + 3\): \(a = -2,\ b = -4,\ c = 3\). \(b^{2} - 4ac = (-4)^{2} - 4(-2)(3) = 16 + 24 = 40\). The tempting \(-8\) takes \(a = 2\), dropping the minus sign on \(x^{2}\). \(8\) and \(-40\) both square \(-4\) to get \(-16\); a square is never negative.
\(b^{2} - 4ac = (-30)^{2} - 4(9)(25) = 900 - 900 = 0\), so there is one repeated root. Indeed \(9x^{2} - 30x + 25 = (3x - 5)^{2}\), so \(x = \dfrac{5}{3}\). The tempting \(-\dfrac{5}{3}\) loses the sign of \(b\): the root is \(-\dfrac{b}{2a} = -\dfrac{-30}{18} = \dfrac{5}{3}\).
\((-k)^{2} - 4(1)(9) = 0\), so \(k^{2} = 36\) and \(k = \pm 6\). Both work: \(x^{2} - 6x + 9 = (x - 3)^{2}\) and \(x^{2} + 6x + 9 = (x + 3)^{2}\). The tempting \(k = 6\) only forgets the negative square root. \(\pm 3\) square-roots 9 instead of 36, and \(k = 36\) stops at \(k^{2}\).
\(36 - 4(3)k < 0\), so \(36 < 12k\) and \(k > 3\). The tempting \(k < 3\) is the condition for two distinct real roots. \(k \ge 3\) wrongly includes \(k = 3\), which gives a repeated root \(x = -1\), and \(k > 12\) uses \(36 - 3k < 0\), forgetting the 4 in \(4ac\).
\(k^{2} - 4(2k - 3) > 0\), so \(k^{2} - 8k + 12 > 0\), i.e. \((k - 2)(k - 6) > 0\). The parabola in \(k\) opens upwards, so it is positive OUTSIDE the roots: \(k < 2\) or \(k > 6\). The tempting \(2 < k < 6\) is the region where the discriminant is negative (no real roots). Using \(\le\) and \(\ge\) includes the equal-root cases, and \(-6\), \(-2\) come from factorising with the wrong signs.
\(x^{2} + 2x + 2 = mx - 2\) gives \(x^{2} + (2 - m)x + 4 = 0\). A tangent meets the curve at one point, so \((2 - m)^{2} - 16 = 0\), giving \(2 - m = \pm 4\): \(m = -2\) or \(m = 6\). The tempting \(-6\) or \(2\) comes from a sign slip when collecting terms, giving \(b = 2 + m\) and \(2 + m = \pm 4\). \(m = 6\) only drops the negative root, and \(2 \pm 2\sqrt{2}\) forgets to move the \(-2\) across, using \(x^{2} + (2 - m)x + 2 = 0\).
Opening downwards needs \(k < 0\). Never meeting the axis needs \(16 - 4k^{2} < 0\), so \(k^{2} > 4\): \(k < -2\) or \(k > 2\). Both conditions together give \(k < -2\). The tempting \(k < -2\) or \(k > 2\) uses only the discriminant: for \(k = 3\) the expression is always POSITIVE. \(k > 2\) is the condition for always positive, and \(-2 < k < 0\) gives a downward curve that does cross the axis.
\(b^{2} - 4ac = (p + 2)^{2} - 4p = p^{2} + 4p + 4 - 4p = p^{2} + 4\). Since \(p^{2} \ge 0\), this is at least 4, always positive, so there are two distinct real roots for every \(p\). The tempting \((p - 2)^{2}\) comes from expanding \((p + 2)^{2}\) as \(p^{2} + 4\) and then subtracting \(4p\). When \(p = 0\) the discriminant is 4, not 0, and an expression that depends on \(p\) can still be positive for every \(p\).
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.