Analysis only. The guide prints the 3kx² + 2x + k question itself, so it is worth treating as the examined case rather than as an extension.
Sweep it slowly from k = 0.9 down past zero and stop on the way. Going down, the order is worth getting right, because it is the opposite of what people expect: at k = 0.9 there are no real roots, the parabola sits clear of the axis, and as k falls the curve widens until at k = 0.577 it just touches, a double root appearing, not two roots merging. Below that it splits into two crossings that keep separating. And as k passes zero the curve straightens into a line, which is where it turns over: the parabola opens upwards for k > 0 and downwards for k < 0, so k = 0 is the flip, not 0.577. Sweeping the other way, upwards towards 0.577, is the one that shows two roots closing up and meeting.
Stop there and read the two numbers out: the discriminant says 4, and there is one root. Ask how both can be true. Because b² − 4ac is a statement about a quadratic, and there is no quadratic. That sentence is the lesson and it is worth getting the class to say it rather than hearing it.
The Δ bar is drawn below the axis when the discriminant is negative, which makes "no real roots" a direction rather than a sign to interpret. Worth pointing at once on the way through k = 1.
| Question | Answer |
|---|---|
| 1. Δ for x² − 4x + 4 | 16 − 16 = 0, so two equal roots, both 2. |
| 2. Real roots of x² + x + 1 | Δ = −3, so none. |
| 3. Positive k for equal roots | k² = 1/3, so k = √3/3 = 0.577. |
| 4. Roots when k = 0 | 2x = 0, so one. |
| 5. Why k = 0 is excluded | B. No x² term, so not a quadratic, so one root rather than two. |
Question 3 wants the surd on Paper 1. √3/3 and 1/√3 are the same number and both are accepted; a decimal is not. Worth insisting on the rationalised form once so they recognise it in a markscheme.
1 marka, b and c identified, with any condition on a stated.
1 markThe discriminant in terms of the unknown.
1 markThe inequality or equation solved.
1 markThe answer as a set of values, exact.
The first mark is the whole reason this page exists. Markschemes award k ≠ 0 explicitly, and it takes seven words. Students who lose it almost never lose it through not knowing; they lose it through not looking at a.
| They wrote | What happened |
|---|---|
| 32 on question 1 | 16 + 16. The formula subtracts, and with c positive the −4ac term is negative. |
| −16 | −4ac with the b² omitted. Usually a student who has written the formula from memory and lost a term. |
| −8 | b² taken as 8 rather than 16, so (−4)² has been read as −4 × 2 or similar. Watch for squaring a negative. |
| 2 on question 1 | The repeated root, not the discriminant. Both are worth having on the board together. |
| 1 on question 2 | Δ = 0 misread as "no roots". Here Δ is actually −3, so they may have conflated the two negative-sounding cases. The geometry fixes it: touching against missing. |
| 0.333 on question 3 | Stopped at k². One square root from correct. |
| 1.732 | √3 rather than √3/3. They have rationalised the wrong way up. |
| 0.667 | 2/3, from rearranging 4 = 12k² carelessly. It is 4/12. |
| 2 on question 4 | The error the whole page is for. They trusted Δ = 4 and did not substitute k = 0 into the equation. Make them substitute. |
"Does Δ = 0 mean one root or two?" Two equal roots, which is one value. IB markschemes say "two equal real roots" and so should they, because a question asking for the number of DISTINCT roots wants one and a question asking about the nature of the roots wants the phrase. Teaching the phrase saves an argument later.
"Why does the curve touching the axis mean equal roots?" Because the two crossings have slid together. Sweep the figure through k = 0.577 slowly and the two marked roots merge into one point. That is the same fact as the ±√0 adding nothing in the formula, and seeing it once means they never have to recall which case is which.
"Can I just graph it?" On Paper 2, to check. Not as the method, because a discriminant question nearly always has an unknown in the coefficients and there is nothing to graph until you have chosen a value. That is worth saying explicitly, since the instinct in a graphing-heavy course is to draw first.
"Is the discriminant ever useful on its own?" Yes, and this is the honest sell: it answers "how many roots" without finding any, which is exactly what questions about tangency, intersection and existence ask. The link forward is 2.15 and AHL 2.12, where the same test decides whether a line meets a curve.
| Demonstrate | Define d(a,b,c)=b^2-4*a*c once, then run the three cases on consecutive lines: 1, 0, −3. Then type d(3k,2,k) and let the machine return 4 − 12k². That last step is the one worth the time, because it shows the discriminant is an expression to be solved rather than a number to be read. Use the x² key or the right arrow to leave the exponent: typing ^ opens a superscript box and everything after it stays inside. |
|---|---|
| Where they stick | The Casio's Polynomial solver reports complex roots or refuses, depending on Complex Mode in SET UP, and students read the refusal as a broken calculator. Set it to Real for this sub-topic so "no solution" means what the question means. On the Nspire, nSolve returns one root and students assume that is all there are; use Analyze Graph → Zero twice instead. |
| The check | Substitute. If k = 0.577 is claimed to give equal roots, put it in and confirm the discriminant comes to zero. If an interval is claimed, test one value inside it and one outside. Two substitutions, and they catch a sign slip in the inequality that no amount of rereading will. |
Analysis Paper 1 has no calculator and this is Paper 1 territory. Set a batch of discriminant questions with the machines face down, then one Paper 2 question where a numerical root is wanted.
| Step | What |
|---|---|
| 1 | Three quadratics on the board, one of each kind. Solve all three by whatever works. |
| 2 | Ask what the three had in common and what differed. Steer to the bit under the root. |
| 3 | Name Δ, and the three cases, with the geometry beside each. |
| 4 | Sweep the figure UP through k = 0.577 so the two roots close up and merge; sweeping down shows the same instant as a double root appearing. |
| 5 | The guide's question: Δ = 4 − 12k², then the interval. |
| 6 | Now sweep through k = 0. Let the curve straighten. Collect the explanation. |
| 7 | Substitute k = 0 by hand: 2x = 0. One root. |
| 8 | Completing the square for an exact answer, to connect back to 2.6. |
Do not say "a negative discriminant means no solutions". It means no REAL solutions, and Higher Level students meet the complex ones at AHL 1.12 and 1.14. Saying it unqualified means a class has to unlearn it a year later, and Standard Level students lose nothing by hearing the word real.
Do not say "find the discriminant first". First comes identifying a, b and c, and noticing whether a can be zero. Put the discriminant second and the condition on a is a step rather than an afterthought, which is the difference between four marks and three.