Date | November 2016 | Marks available | 6 | Reference code | 16N.1.AHL.TZ0.H_5 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | H_5 | Adapted from | N/A |
Question
The quadratic equation x2−2kx+(k−1)=0 has roots α and β such that α2+β2=4. Without solving the equation, find the possible values of the real number k.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
α+β=2k A1
αβ=k−1 A1
(α+β)2=4k2⇒α2+β2+2αβ⏟k−1=4k2 (M1)
α2+β2=4k2−2k+2
α2+β2=4⇒4k2−2k−2=0 A1
attempt to solve quadratic (M1)
k=1, −12 A1
[6 marks]
Examiners report
[N/A]
Syllabus sections
Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots
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