Free Quadratics & the discriminant questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Functions. Each question has the answer and the marking points an examiner would look for.
Generate a full Quadratics & the Discriminant worksheet →All Functions questionsThe line y = kx + 5 is a tangent to the curve y = x2 + 6.
Find the possible values of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| k = −2 or k = 2 x2 + 6 = kx + 5 M1 x2 − kx + 1 = 0 A1 discriminant = 0 M1 k2 = 4 (A1) k = −2 or k = 2 A1 | 5 |
Consider the equation x2 + kx + 2k = 0, where k ∈ ℝ.
Find the set of values of k for which the equation has two distinct real roots.
| Part | Answer and marking points | Marks |
|---|---|---|
| k < 0 or k > 8 discriminant > 0 M1 (k)2 − 4(2k) > 0 A1 k2 − 8k > 0 A1 critical values k = 0, 8 A1 correct region chosen (outside the roots) (M1) k < 0 or k > 8 A1 | 6 |
The line y = kx is a tangent to the curve y = x2 − 4x + 1.
Find the possible values of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| k = −6 or k = −2 x2 − 4x + 1 = kx M1 x2 + (−4 − k)x + 1 = 0 A1 discriminant = 0 M1 (−4 − k)2 = 4 (A1) k = −6 or k = −2 A1 | 5 |
Consider the equation x2 + (k + 8)x + 2k + 13 = 0, where k ∈ ℝ.
Find the set of values of k for which the equation has two distinct real roots.
| Part | Answer and marking points | Marks |
|---|---|---|
| k < −6 or k > −2 discriminant > 0 M1 (k + 8)2 − 4(2k + 13) > 0 A1 k2 + 8k + 12 > 0 A1 critical values k = −6, −2 A1 correct region chosen (outside the roots) (M1) k < −6 or k > −2 A1 | 6 |
The line y = kx − 2 is a tangent to the curve y = x2 − 2x − 1.
Find the possible values of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| k = −4 or k = 0 x2 − 2x − 1 = kx − 2 M1 x2 + (−2 − k)x + 1 = 0 A1 discriminant = 0 M1 (−2 − k)2 = 4 (A1) k = −4 or k = 0 A1 | 5 |
Consider the equation x2 + (k − 4)x − 2k + 5 = 0, where k ∈ ℝ.
Find the set of values of k for which the equation has two distinct real roots.
| Part | Answer and marking points | Marks |
|---|---|---|
| k < −2 or k > 2 discriminant > 0 M1 (k − 4)2 − 4(−2k + 5) > 0 A1 k2 − 4 > 0 A1 critical values k = −2, 2 A1 correct region chosen (outside the roots) (M1) k < −2 or k > 2 A1 | 6 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in