I was doing some basic algebra with sage and encountered another
curiosity. I wish to solve for x in:
|x^2-x| = 3
My naive attempt in sage wasn't so fruitful
sage: eqn = maxima('abs(x^2-x)=3')
sage: eqn.solve('x')
[abs(x^2 - x) = 3]
Although, the subproblems are easy enough:
sage: eqn = maxima('x^2-x=-3')
sage: eqn.solve('x')
[x = - (sqrt(11)*%i - 1)/2,x = (sqrt(11)*%i + 1)/2]
sage: eqn = maxima('x^2-x=3')
sage: eqn.solve('x')
[x = - (sqrt(13) - 1)/2,x = (sqrt(13) + 1)/2]
How should such a difficulty be attacked? Is this a deficiency of
maxima?
-carson-
PS: Thanks for the helpful discussion of methods for computing the
conjugate tranpose of matrices.
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