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    This relates to Question 3 of STEP I 2007, which can be found at
    https://3110b3b48cdc993f8fb622042e2f...20STEP%201.pdf.
    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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    (Original post by HapaxOromenon3)
    This seems much shorter and quicker than the process which the question attempts to guide you through, so would it score all the marks?
    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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    (Original post by Zacken)
    No, of course not. You've skipped a good bit of the question and only answered the easy "deduce" bit at the end...
    I think he means for that prt.


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    (Original post by physicsmaths)
    I think he means for that prt.


    Posted from TSR Mobile
    I don't think so, look at his post:

    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"
    Had he meant only the deduce part, then he'd have said "which starts "deduce that..."".
 
 
 
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