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?
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- Thread Starter
- 17-07-2016 06:41
- 17-07-2016 10:51
(Original post by Zacken)
- 17-07-2016 12:51
No, of course not. You've skipped a good bit of the question and only answered the easy "deduce" bit at the end...
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- 17-07-2016 14:19
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"