Re: quadratic solution

"Phil Smith" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Today Ken Oliver wrote in sci.math:
>
> > A colleague from my pre-retirement workplace asked me the following:
> >
> > Solve:
> >
> >         (5 + sqrt(x))^2 - 9*(5 + sqrt(x) ) + 20 = 0
> >
> > This was a test question she had used previously asking for a solution
over
> > the reals.  No problem.
> >
> > But this year she put this on a test covering quadratics over the
complex
> > #'s with coeff's possibly in C.
> >
> > By the way, sqrt() is traditional "check mark-overscore thingy."
> >
> > This led to needing a solution of sqrt(x) = -1.  A root of course was
> > discarded on the real number test, but what about here.
> >
> > What is the solution to this over the complex numbers?  "What the rules
are
> > for applying sqrt() to complex radicand" actually seems to be what the
> > problem comes down to.
> >
> > I a little embarassed to say that I don't know the rules for "principle
> > values" or branching here.
>
> Depends how you define sqrt(z).
>
> If you make a cut along the +ve real axis
> so 0 < arg(z) <= 2*pi and define sqrt(1.exp(i*0)) = 1, then
>
> sqrt(z) = -1 => sqrt(z) = exp(i*pi) => z = exp(2*pi*i) - the other side of
> the cut.
>
> Alternatively, you could take -pi < arg(z) <= pi and sqrt(1) = 1; then
> sqrt(z) = -1 => arg(z) = 2*pi; no solution.
>
> -- 
> P.A.C. Smith
> replying by email: s/NOSPAM//
>
> "The vast majority of Iraqis want to live in a peaceful, free world.
> And we will find these people and we will bring them to justice."
>  - George W. Bush (Washington DC, Oct 27 2003)
>

Let u=5+sqrt(x)
u^2-9u+20=0
(u-4)(u-5)=0
u=4 or u=5

4=5+sqrt(x)
No solution

5=5+sqrt(x)
x=0

David Moran