[Blindmath] Factoring Question.

Jonathon Yaggie jyaggi2 at uic.edu
Thu Oct 24 21:15:44 UTC 2013


if it is not a typo.  then the roots are the roots of x^2=-1-\sqrt{2} and
x^2=-1+\sqrt{2} .  the first quadratic has imaginary solutions and the
second has real solutions.  some one with a calculator could find the four
roots - two imaginary and two real.


On Thu, Oct 24, 2013 at 4:06 PM, Tami Jarvis <tami at poodlemutt.com> wrote:

> Yeah, I keep thinking there may be something involving the imaginary
> number i (the positive or negative square root of 1). It has been more
> years than I care to admit, though, so I'm not getting anywhere trying to
> use it to step through the equation to see if it works...
>
> The assumption that the equation is a typo makes it really easy, and it
> all works out nice and neat. :)
>
> Tami
>
>
> On 10/23/2013 09:21 PM, I. C. Bray wrote:
>
>>          Sorry, I misplaced / deleted the post I was replying to due to a
>> phone call that interrupted my posting.
>>
>> To whom it was that asked the factoring question.
>>
>> That expression is unfactorable over the set of Real numbers.
>> it has roots of:
>> "Plus or Minus" the square root of (1+the square root of(2) )
>>
>> That is an Irrational number due to the square root of 2...
>>
>> you can say that it is Prime with respect to real numbers
>> or that it's roots are irrational, and I believe even imaginary.
>>
>>
>>
>> Respectfully,
>>
>> Ian  C. Bray
>>
>>
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>>
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-- 
Jon Yaggie
UIC Mathematics



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