IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~7 min read

Discriminants

The discriminant is the bit under the square root in the quadratic formula: b² − 4ac. Its sign alone tells you how many real roots a quadratic has — without solving anything. Most exam questions in this topic don’t ask you to find roots; they ask you to find a value or range of an unknown letter (usually k) that gives a specific number of roots. The discriminant turns that into one neat equation or inequality.

📘 What you need to know

Discriminant: Δ = b² − 4ac. In the formula booklet.
Δ > 0 ⟹ two distinct real roots (graph cuts the x-axis at two points).
Δ = 0 ⟹ one repeated real root (graph touches the x-axis at the vertex).
Δ < 0 ⟹ no real roots (graph doesn’t touch the x-axis at all).
Δ ≥ 0 ⟹ “real roots” — covers both 1 and 2 (use this when the question doesn’t specify how many).
For tangent conditions (a line just touches a curve once), set up the intersection equation and require Δ = 0.
If a quadratic has unknown leading coefficient k, also remember k ≠ 0 — otherwise it isn’t quadratic at all.

The three cases — what Δ tells you

Δ > 0

2 distinct real roots

graph crosses the x-axis at two points

Δ = 0

1 repeated root

graph just touches the x-axis (tangent)

Δ < 0

no real roots

graph never reaches the x-axis

Three parabolas, three discriminant cases

If a question says “real roots” without specifying how many, that’s the inclusive case — use Δ ≥ 0. “Distinct real roots” means strict — use Δ > 0. Read the wording carefully; one missing word changes the answer.

The standard k-question recipe

Most discriminant questions follow the same shape: you’re given a quadratic with an unknown k, told something about the roots, and asked to find k. Here’s the routine:

🧭 Recipe — discriminant questions with a parameter

Identify a, b, c in terms of k. Watch signs.
Compute Δ = b² − 4ac as an expression in k.
Apply the right condition: Δ > 0 (2 distinct), Δ = 0 (repeated), Δ < 0 (none), Δ ≥ 0 (real).
Solve the resulting equation or inequality in k.
Check k ≠ 0 if k appears as the leading coefficient (otherwise the equation isn’t quadratic).

Tangent conditions — Δ = 0 in disguise

When a question says “the line is a tangent to the curve” or “the line touches the curve at exactly one point”, the meeting equation has exactly one solution — so its discriminant is zero. Set up the intersection, rearrange to a quadratic in x, and apply Δ = 0.

Worked examples

WE 1

Classify the number of roots

For each equation, find the discriminant and state the number of real roots:
(a) x² − 6x + 9 = 0 (b) 2x² + 3x − 1 = 0 (c) x² + x + 5 = 0

(a) a = 1, b = −6, c = 9 Δ = 36 − 36 = 0 → one repeated root (b) a = 2, b = 3, c = −1 Δ = 9 − 4(2)(−1) = 9 + 8 = 17 → two distinct real roots (c) a = 1, b = 1, c = 5 Δ = 1 − 20 = −19 → no real roots (a) 1 repeated (b) 2 distinct (c) 0 real when c is negative, −4ac becomes positive — easy place to slip up

WE 2

Show a quadratic has two distinct real roots

Show that the equation 3x² − 7x + 2 = 0 has two distinct real roots.

Identify a, b, c a = 3, b = −7, c = 2 Compute Δ Δ = (−7)² − 4(3)(2) = 49 − 24 = 25 Δ = 25 > 0, so the equation has two distinct real roots ✓ since Δ is a perfect square the roots are also rational — they would factorise nicely

WE 3

Find k for a repeated root

The equation x² + (k + 2)x + 9 = 0 has a repeated real root. Find the possible values of k.

Step 1: a = 1, b = k + 2, c = 9 Step 2: Repeated root → Δ = 0 (k + 2)² − 4(1)(9) = 0 (k + 2)² − 36 = 0 (k + 2)² = 36 Step 3: ±√ both sides k + 2 = ±6 k = 4 or k = −8 k = 4 or k = −8 don’t forget the negative root — both values give a repeated solution

WE 4

Find a range of k for two distinct real roots

The equation kx² + 4x + (k − 3) = 0 has two distinct real roots. Find the set of values of k.

Step 1: a = k, b = 4, c = k − 3 — and need k ≠ 0 Step 2: Two distinct roots → Δ > 0 16 − 4(k)(k − 3) > 0 16 − 4k² + 12k > 0 divide by −4 and FLIP: k² − 3k − 4 < 0 Step 3: Solve the inequality (k − 4)(k + 1) < 0 roots −1, 4 → between (∪-shape, want < 0) −1 < k < 4 Step 4: Exclude k = 0 −1 < k < 4, k ≠ 0 when k appears as the leading coefficient, always exclude k = 0 — at that value the equation is linear, not quadratic

WE 5

Tangent line to a parabola

The line y = kx + 2 is a tangent to the curve y = x² + 5x + 6. Find the possible values of k.

Step 1: Set the two equal — tangent touches at exactly one point x² + 5x + 6 = kx + 2 x² + (5 − k)x + 4 = 0 Step 2: One solution → Δ = 0 (5 − k)² − 4(1)(4) = 0 (5 − k)² = 16 Step 3: ±√ both sides 5 − k = ±4 k = 1 or k = 9 k = 1 or k = 9 two tangent lines exist with this y-intercept — one with gentle slope, one steep, each touching at a different point

WE 6

Find a range of p for no real roots

The equation x² + 2px + (p + 6) = 0 has no real roots. Find the set of values of p.

Step 1: a = 1, b = 2p, c = p + 6 Step 2: No real roots → Δ < 0 (2p)² − 4(1)(p + 6) < 0 4p² − 4p − 24 < 0 divide by 4 (positive — sign stays): p² − p − 6 < 0 Step 3: Solve the inequality (p − 3)(p + 2) < 0 roots −2, 3 → between −2 < p < 3 no need to exclude p = 0 here because the leading coefficient is 1, not p

💡 Top tips

Memorise the three cases. Δ > 0 (two distinct), Δ = 0 (repeated/touches), Δ < 0 (none).
“Real roots” without “distinct” means Δ ≥ 0. “Distinct real roots” means Δ > 0. Check the wording.
Tangent = Δ = 0. Whenever a line “just touches” a curve, set up the intersection and force the discriminant to zero.
Always check k ≠ 0 when k is the leading coefficient. Don’t lose marks on this technicality.
Compute Δ on its own line first — easier to spot sign errors than working it inside the formula.
Don’t forget the ± when square-rooting (k + something)² = number. Two answers, not one.
For Δ > 0 inequalities, use the standard quadratic-inequality recipe — sketch and pick the region.

⚠ Common mistakes

Confusing Δ > 0 with Δ ≥ 0. “Distinct” excludes the equal case; without “distinct” it includes it.
Sign error on −4ac when c is negative. Two negatives multiply to a positive: −4(2)(−5) = +40, not −40.
Forgetting k ≠ 0 when the leading coefficient is k. Without that, the equation could be linear.
Forgetting to flip the inequality when dividing by a negative coefficient on the way to solving Δ < 0 or Δ > 0.
Treating “tangent” as “intersects”. A tangent meets the curve at exactly one point — that’s Δ = 0, not Δ ≥ 0.
Missing the ± when square-rooting a perfect square. (k − 1)² = 9 gives k − 1 = ±3, so k = 4 or k = −2.
Using Δ on something that isn’t quadratic. If the highest power isn’t x², the discriminant rule doesn’t apply.

And that’s the end of Section 2.2 — Quadratic Functions. You’ve now got the full toolkit: graph features, factorising, completing the square, three solving methods, inequalities, and discriminants. These come up again and again throughout the rest of the syllabus — domains and ranges, tangent problems in calculus, optimisation, even probability with quadratics in disguise. The next section, 2.3, zooms out to functions in general — domain, range, function notation, and what makes something a function in the first place.

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