Re: JSH: Gotcha

In sci.math, jstevh@xxxxxxxxx
<jstevh@xxxxxxxxx>
wrote
on 28 Apr 2007 21:43:42 -0700
<1177821822.884887.303000@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick--multiply the last two
coefficients of the original polynomial with one solution for x and
I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 +
sqrt(-26) as a factor,

If you're going to do that, do it properly. If one has the
polynomial equation x^2 + Ax + B = 0, one can substitute
y = x/C and then multiply by C^2, yielding
y^2 + A*C*y + B*C^2 = 0. It's easily proven that the latter has
roots y1 = x1 * C, y2 = x2 * C, if the original equation had
roots x1 and x2.

Applying this to your equations above, one gets:

x^2 - 6(3+sqrt(-26))x + 35(9 - 26 + 6*sqrt(-26)) = 0

One can now guarantee that the roots of this equation
have the factor 3 + sqrt(-26). The one root, of course,
is -17 + 6 * sqrt(-26). The other is 35. Clearly the latter
is divisible by 7.

With your original suggestion of

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

one gets the rather ugly but straightforward expression
from the standard quadratic solution:

x1 = (-18-6*sqrt(-26) + sqrt(324-936-36*2*sqrt(-26) - 4*35*(3 + sqrt(-26))))/2
= -9-3*sqrt(-26) + sqrt(81-234-9*2*sqrt(-26) - 35*(3 + sqrt(-26)))
= -9-3*sqrt(-26) + sqrt(-153-18*sqrt(-26) - 105 + 35*sqrt(-26)))
= -9-3*sqrt(-26) + sqrt(-258+17*sqrt(-26))

The other root is, of course,

x2 = -9-3*sqrt(-26) - sqrt(-258+17*sqrt(-26))

Neither of these appear divisible by 7. Were I to grind through it I'd
probably get a quartic, if not an octic.

but notice that one root must have 7 as a
factor as well, as the last coefficient has 7 as a factor, so this is
just a tricky way of forcing one root to have 7 as a factor without
forcing them both.

Now I can group my radicals to one side and everything else on the
other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

And now you're into extremely problematic territory.

Let x = 1.
Then x^2 = 1.
Therefore, x = 1 or -1.
Therefore, -1 = 1. QED.

This is oversimplified but variants of this problem
occasionally trip up those who want to solve equations
with radicals such as the one you posed above.


which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor.

But that polynomial is irreducible over Q.

So is x^2 - 35. Do the roots thereof have either 5 or 7 as a factor?


So you see, I've been right all along and you people have been
fighting very valuable and important mathematics.

You've been fighting against the discipline of mathematics itself in a
futile attempt to destroy the future of humanity.

You have been fighting history.

Your Nobel Prize is in the mail. :-)



James Harris


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#191, ewill3@xxxxxxxxxxxxx
If your CPU can't stand the heat, get another fan.

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