Date | November 2021 | Marks available | 5 | Reference code | 21N.1.AHL.TZ0.7 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The equation 3px2+2px+1=p has two real, distinct roots.
Find the possible values for p.
Consider the case when p=4. The roots of the equation can be expressed in the form x=a±√136, where a∈ℤ. Find the value of a.
Markscheme
attempt to use discriminant b2-4ac(>0) M1
(2p)2-4(3p)(1-p)(>0)
16p2-12p(>0) (A1)
p(4p-3)(>0)
attempt to find critical values (p=0, p=34) M1
recognition that discriminant >0 (M1)
p<0 or p>34 A1
Note: Condone ‘or’ replaced with ‘and’, a comma, or no separator
[5 marks]
p=4⇒12x2+8x-3=0
valid attempt to use x=-b±√b2-4ac2a (or equivalent) M1
x=-8±√20824
x=-2±√136
a=-2 A1
[2 marks]