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Spanish Problem and solving x3+px=q
By Adrian on March 18, 1998:

This query refers to a question set several years ago in the Spanish Universities entrance examination (`selectividad') which translated reads as follows:


Find the point on the graph of the function y=x2-1 that is closest to the point (2,-1).


It appears to be a fairly straightforward case of minimising a general formula for distance between the given point and the given graph but this leads to having to solve a cubic equation which cannot be solved by normal `A level' methods. This leads us to suspect that there may be an error in the choice of function or co-ordinates. Do you agree or are we missing something obvious?

By Rup on March 20, 1998:

It's not nice, is it? I get 2x3+x-2=0 from your suggested distance minimisation and have verified this by finding the normal to the curve.

You could always quote the result by Cardon -- maybe this is on their syllabus. One of my A-level teachers wrote it on the board to get himself out of a similar problem but I haven't seen it since. Took a bit of finding, too :-)

General solution to x3+px=q

x= cubert( sqrt((q/2)2+(p/3)3) + (q/2) )

-cubert( sqrt((q/2)2+(p/3)3) - (q/2) )


which, for p=1/2, q=1, gives:

x=cubert(sqrt(330)/36 + 1/2)-cubert(sqrt(330)/36 - 1/2)


However, quoting a result isn't terribly taxing; on the other hand, if the question was wrong and it /was/ solvable by A-level means.... that wouldn't be terribly taxing either.


There is also a method of solving such things using a sin substitution and De Moivre to express sin3 x in terms of sin 3x but I'm afraid I'm not familiar with it. I also don't under what conditions Cardon holds or how it stands up to equations with three solutions instead of just one; applying it to x3-7x=-6 (solutions 1,2,-3) gives a complex cube root of one.

Hope this helps (in some small way, at least!)

Rup.