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And for nothing, I agreed.
Not for nothing. I have an endless family of solutions in the back of my mind.
And, by the way, a cubic equation always has at least one valid root.
Not for nothing. I have an endless family of solutions in my stash.
Oh, by the way, a cubic equation always has at least one valid root.
Where did it go?
Is the calculator lying?
// Solved here http://web2.0calc.com/
Showing.
X
k*X
k^2*X + N(X + k*X)
Is the calculator lying?
Where did it go?
Is the calculator lying?
Looks like it's lying. If it's solving numerically, it probably overflows.
(sighs) I don't know.
And by the way, a cubic equation always has at least one valid root.
Is it not for equations of the form ax^3+bx+c=0?
?
Anything can happen when x^2 appears...
No, it can't. It turns out all cubic equations are reducible to the form x^3+px+q=0.
No, it can't. It turns out that all cubic equations are reducible to the form x^3+px+q=0
Very easy to justify logically. infinite minus at minus infinity, plus the opposite, so the x-axis is crossed at least once, since the function is continuous.
I have a general suspicion that all equations in question have all three valid roots, of which one is positive. The degrees at i in your screenshot confirm it.
It is very easy to justify logically. infinite minus at minus infinity, plus vice versa, so the x-axis is crossed at least once, since the function is continuous.
I have a general suspicion that all equations in question have all three valid roots, of which one is positive. The degrees at i in your screenshot confirm it.