(In reply to
go graphical by Larry)
Thank you for your excellent observations, Larry, which have led me to the following approach.
I f we let u=x-2, the equation becomes
sqrt(1+u) + sqrt(1-u) = u^2 + 2.
The equation is now symmetrical around u=0, and clearly has u=0 as a
solution. Thus we need only consider u>0 in seeking other
possible solutions. And we need not consider u>1 as sqrt(1-u) is not
real when u>1. The derivatives of the lhs and rhs are equal when u=
0, but as u increases, they have opposite sign since clearly the
derivative of the rhs is positive, and that of the lhs is
(1/2)(1/sqrt(1+u) - 1/sqrt(1-u))
and 1/sqrt(1-u) > 1/sqrt(1+u). Hence the lhs and rhs diverge away
from each other as u increases from 0 to 1. u=0 is the only
solution.
Edited on August 17, 2006, 12:37 am
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Posted by Richard
on 2006-08-17 00:33:03 |
re: go graphical
