This relates to Question 3 of STEP I 2007, which can be found at
I have an alternative method that is not listed in the Solutions or Examiner's Report, for the part of the question which starts "It is given that the equations ay^2+by+c=0 and by^2+cy+a=0 have a common root k". I would like to know whether it would be considered creditworthy.
My solution is as follows: write ak^2 + bk + c = 0 and bk^2 + ck + a =0.
If k = 0 then the second equation gives a = 0, but we are told that a is non-zero, so k must be non-zero. Thus we can divide the second equation by k to give
bk + c + a/k = 0. Now subtract this from the first equation to give ak^2 - a/k = 0, which becomes k^3 - 1 = 0 -> (k-1)(k^2+k+1)=0, so either k = 1 or k^2 + k + 1 = 0. The second case then implies that ak^2 + ak + a = 0 and bk^2 + bk + b = 0, so that a = b = c (to be consistent with the original equations), hence the equations are identical. Thus we have shown the required result: either k = 1 or the equations are identical.
This seems much shorter and quicker than the process which the question attempts to guide you through, so would it score all the marks?
STEP Question Watch
- Thread Starter
- 17-07-2016 07:41
- 17-07-2016 11:51
(Original post by Zacken)
- 17-07-2016 13:51
No, of course not. You've skipped a good bit of the question and only answered the easy "deduce" bit at the end...
Posted from TSR Mobile
- 17-07-2016 15:19
I have an alternative method that is not listed in the Solutions or Examiner's Report, for the part of the question which starts "It is given that the equations ay^2+by+c=0 and by^2+cy+a=0 have a common root k"