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Solving an equation - enzor - 01-27-2021 01:28 PM
Hi, i was solving the next equation:
solve(abs((kc+g/u)/(1+kc))=1,kc)
and the solution I get is:
(2*i*_y*u-g-u)/(2*u)
Does anyone knows what _y means?
Thanks in advance!
RE: Solving an equation - Albert Chan - 01-27-2021 05:44 PM
(01-27-2021 01:28 PM)enzor Wrote: solve(abs((kc+g/u)/(1+kc))=1,kc)
and the solution I get is:
(2*i*_y*u-g-u)/(2*u)
Does anyone knows what _y means?
_y is just a notation for any constant. (note: above solution assumed g/u is real).
Let x=kc, g/u=m:
If x≠-1, |x+m| = |x+1|
Square both side, and simplify:
x² + 2mx + m² = x² + 2x + 1
2*(m-1)*x = 1-m² = -(m+1)*(m-1)
x = -(m+1)/2
Above solution, for x=-1 -> m=1 -> above solution work for x=-1 too.
If we assume m is real -> x is real.
We can add imaginary part to x, and still satisfy |x+m| = |x+1|
→ kc = x + _y*i = -(g/u+1)/2 + _y*i