At a glance. One lesson of about 60 minutes. It assumes factorizing (Lesson 2) and completing the square (Lesson 3). Students derive the formula, use the discriminant to count and classify solutions, choose a method, and find a parameter.
Calculator: GDC optional. Everything can be done by hand. The optional section shows the zero tool and the polynomial solver, and warns that the exact form needs the working.
What students will learn
Know
- The quadratic formula and the discriminant Δ = b² − 4ac.
- Δ > 0: two solutions; Δ = 0: one; Δ < 0: none.
- A perfect-square Δ means rational solutions.
Understand
- The formula is completing the square done once in general.
- The sign of Δ says where the parabola is relative to the axis.
- The best method depends on the numbers.
Do
- Derive the formula by completing the square.
- Solve by the most efficient method.
- Find a parameter for a given number of solutions.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: solving and using the discriminant (questions 1 to 7).
- Criterion B. Investigating patterns: the what-if and the figure. Students find where the number of solutions changes and explain it with the vertex.
- Criterion C. Communicating: the derivation in the key card, and the checks for the equation in disguise (worked example 3).
- Criterion D. Applying mathematics in real-life contexts: the stall's income, deciding if a target is reachable before solving.
The one thing to do with the figure
Ask for a prediction before every slide. With b = 0 on the screen, ask: “At what value of b will the parabola first touch the axis?” Then move b slowly and let students watch the ring on the Δ graph cross the axis.
Then ask why the right-hand graph is itself a parabola in b, and where its zeros are. Students should see that those are the b values for a repeated solution.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the stall. Collect predictions about the number of solutions. |
| 12 min | The figure with predictions; the key idea and the derivation; settle the hook, then answer the stall's question with Δ. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose B or C, return to Δ and the inequality. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | x = −3 or 8; −8; 5√3. |
|---|---|
| Hook | Yes: the number of solutions changes where Δ = 0, at b = 6 and b = −6. For the stall, x(20 − x) = 100 gives x² − 20x + 100 = 0, with Δ = 400 − 400 = 0: exactly one amount, 10 kg. A target of 120 gives Δ = 400 − 480 < 0, so it can never be reached. |
| Worked examples | x = −1/3 or 2; (−3 ± √11) ÷ 2 (about 0.16 and −3.16); x = 2 or 7. |
| Faded example | k > 25. |
| 1. A parameter and no real solutions | A. B flips the inequality; C uses 2ac; D uses ac. |
| 2. Not zero on one side | 4. (x + 7)(x − 4) = 0. |
| 3. Evaluating the discriminant | 24. 64 − 40. |
| 4. Using the formula | 5. (7 + 13) ÷ 4. |
| 5. Repeated solution | 8. b² = 64. |
| 6. A parameter for two solutions | 3. 36 − 12k > 0. |
| 7. An equation in disguise | 3. x = 3 or 4. |
| 8. Counting solutions with the discriminant | C. Δ = 0; A has Δ < 0; B and D have Δ > 0. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| B, C or D in question 1 | Flipped the inequality, or used the wrong multiple of ac. | Write Δ = b² − 4ac with a and c named. |
| 20 or 13 in question 2 | Set a bracket equal to 18. | Ask: what does the null factor law need on the right side? |
| 104 or 44 in question 3 | Added the 4ac term, or left a out. | Write 4ac = 4 × 2 × 5 on its own line. |
| 10, 1.5 or −1.5 in question 4 | Divided by 2, subtracted the wrong way, or gave the other solution. | Write 2a = 4, then both solutions. |
| 4, 64 or 32 in question 5 | Gave the root, b², or doubled the constant. | Set Δ = 0 and solve for b. |
| 9 or 6 in question 6 | Left out c, or used 2ac. | Write 4ac with the numbers in. |
| 4 or 7 in question 7 | Gave the larger solution or the sum. | List both solutions, then choose. |
| A, B or D in question 8 | Counted a Δ that was not 0. | Work out Δ for each option. |
Questions worth asking
- Why does Δ < 0 mean the parabola misses the x-axis? When Δ < 0, the vertex height k = −Δ ÷ (4a) has the same sign as a: an upward parabola has its vertex above the axis, a downward one below, so it never reaches it.
- When is a repeated solution the vertex? When the vertex lies on the axis, so the two zeros have merged.
- How does Δ tell you whether it factorizes? If Δ is a perfect square the solutions are rational and whole-number brackets exist.
- Which method for x² + 6x − 3 = 0? Not factorizing: Δ = 48 is not a perfect square, so use the formula or complete the square.
Differentiation
A way in for anyone stuck
- Give a card with Δ = b² − 4ac and a small table of what each sign means.
- Write a, b and c above the equation before substituting.
- Practise with Δ only first, then add the formula.
Stretch
- Show that if a and c have opposite signs, the equation always has two real solutions.
- Find k so that kx² + 4x + k = 0 has equal solutions, and say what the graph looks like.
Students learning in English as an additional language
“Discriminant” is the name for b² − 4ac; it “discriminates” (tells apart) the cases. “Repeated solution” means one solution counted twice. “Real” means an ordinary number on the number line. Say “delta” for Δ.
| Word | Plain meaning | Example |
|---|---|---|
| discriminant | b² − 4ac, the part under the root in the formula | Δ = 49 − 40 = 9 |
| real solution | a solution that is an ordinary number | x = 6 for x − 6 = 0 |
| repeated solution | one solution counted twice, when Δ = 0 | x = −2 for (x + 2)² = 0 |
| perfect square | a whole number that is a whole number squared | 49 = 7² |
| quadratic formula | x = (−b ± √(b² − 4ac)) ÷ (2a) | solves ax² + bx + c = 0 |
| irrational | a number that cannot be written as a fraction | √13 |
Sentence starters for explaining aloud or in writing:
- The discriminant is … so there are …
- I chose to … because the numbers …
- The solutions are exact because …
Two things not to say
“Every quadratic has two solutions.” It can have two, one or none.
“If it does not factorize, it has no solutions.” It may have irrational solutions; check Δ.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- SolveSolve x² − 9x + 20 = 0.
x = 4 or x = 5. - CountHow many real solutions has x² + 2x + 5 = 0?
None: Δ = 4 − 20 = −16.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA rectangular garden in Nairobi has area 66 m² and the length is 5 m more than the width. Find the width.
w(w + 5) = 66 gives w² + 5w − 66 = 0, so (w + 11)(w − 6) = 0 and w = 6 m. - Earlier work (Lesson 3)Write y = x² + 8x + 3 in vertex form.
y = (x + 4)² − 13.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningFind k so that x² + kx + 36 = 0 has exactly one solution.
k² − 144 = 0, so k = ±12. - Method choiceSolve 3x² − 6x − 2 = 0 exactly.
Δ = 36 + 24 = 60, x = (6 ± 2√15) ÷ 6 = (3 ± √15) ÷ 3.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to the slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
The With a GDC section is closed by default and optional. Students without a GDC lose nothing; every answer is reached by hand.