Date | May 2021 | Marks available | 5 | Reference code | 21M.1.AHL.TZ2.7 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The cubic equation x3-kx2+3k=0 where k>0 has roots α, β and α+β.
Given that αβ=-k24, find the value of k.
Markscheme
α+β+α+β=k (A1)
α+β=k2
αβ(α+β)=-3k (A1)
(-k24)(k2)=-3k (-k38=-3k) M1
attempting to solve -k38+3k=0 (or equivalent) for k (M1)
k=2√6 (=√24)(k>0) A1
Note: Award A0 for k=±2√6 (±√24).
[5 marks]
Examiners report
[N/A]