Re: Help (polynomial)


<john_ramsden@xxxxxxxxxxxxxx> ha scritto nel messaggio
news:1117118897.747286.76890@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> big.ass wrote:
>>
>> Given two integers m,n such that m < n, prove that
>> if k > (m^2+n^2)/2 (with k real number) then the
>> polynomial
>>
>> p(x)=(x^2+k)(x-m)(x-n)+1
>>
>> has two real roots and two non-real roots.
>
> I don't think it makes any difference whether m and n are
> integers or rational, as the conditions relate only to real
> numbers.
>
> Denoting p(x) by (x^2 - 2u.x + u^2.v).(x^2 + 2u.x + u^2.w) + 1,
> your condition m^2 + n^2 < 2k is equivalent to 0 < v + w.




> But denoting p(x) by (x^2 - 2a.x + a^2.b).(x^2 + 2a.x + a^2.c),
> the condition for two real and two non-real roots is equivalent
> to (1 - b).(1 - c) < 0.


A bit careful here: I can't assume that p(x) can be written in that form
because it is just what have to prove.



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