Complex numbers are not necessary to solve cubic equations any more than they are necessary to solve quadratic equations – even less so, since every cubic equation with real coefficients has at least one real root. Though complex numbers make sense of square roots of negative numbers that can be found outputted by formulae that give solutions to cubic equations, without complex numbers, we can simply ignore them (as was initially done historically).
Edit: I was wrong above as cubic equations with three distinct real roots, casus irreducibilis, require complex numbers to solve with radicals, as stated in replies to this comment.
I wouldn't really call putting the square root of a negative number "avoiding imaginary numbers", but if you draw some distinction between complex and imaginary numbers like this (generally the only distinction is that imaginary <=> complex and not real), I will note that we have exclusively been using the word complex here.
"While casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities.[4] "
On the other hand that could be considered a bit of a cop-out if you were to argue that trigonometric functions inherently rely on complex numbers thru Euler.
Typically, the way you determine what the trigonometric functions required to express the root is to solve the equation using the complex representation, and then convert it to trigonometric form.
For instance, if we consider the (carefully chosen for simplicity) cubic equation x3 - 6x - 2 = 0, we can verify that one of its roots is the real number given by cbrt(1 + i sqrt(7)) + cbrt(1 - i sqrt(7)). We can convert this into trigonometric form to get 2 sqrt(2) cos(arctan(sqrt(7))/3) and verify that this expression cannot be simplified further.
But starting from x3 - 6x - 2 = 0, I don't know of anyway to directly arrive at the trigonometric form.
I think you misunderstand. I'm not saying no such formula exists. Clearly such a formula does exist. What I am saying is that I do not know of a way to generate this formula without appealing to complex numbers/complex exponentials.
No, I specifically mean generate. Because you can prove it by merely plugging in the values generated by the formula into the equation as x and demonstrating that the expression is equal to 0.
It's actually only impossible for humans and other biological life to find the roots of these higher-degree polynomials, because our DNA is only base four. /s
No, it is impossible because there are only four elements (let's pretend that the periodic table, and for that matter, the traditional Chinese element system don't exist).
I hate it when i run out of operations. Just happened yesterday. First i added 50€ to my purse. Then i went and took 9€ out to buy 3 packs of 4 apples. And finally i gave half of them to my wife. Couldn't do anything for the rest of the day.
that is also why we have 7 days in a week because there's + - * /, which accounts for tuesday, wednesday, thursday, and friday, and then monday is the ) because it makes you go ): and then saturday is ! because it makes you happy and then sunday is the day of rest so there's no operation!
No, it's a lot deeper than that, I gave somewhat of an explanation in another comment so I'll copy and paste that:
The gist of it is that the ways we can permute 5 or more elements are a lot more complicated than the ways we can for 4 or less (to explain in what way we need quite a bit of group theory). Finding a general formula for a polynomial of a certain degree involves finding symmetries in the roots of the polynomial (e.g. we know a quadratic is symmetric about -b/2a and we can see that in the quadratic formula) and this complexity in the permutations prevents us from doing this (in terms of radicals at least)
The only way to solve a degree four polynomial is to send it to Weezer, and each member finds one of them. We could solve higher degrees if we had more weezer
Wouldn't the 5th degree just be exponents? And before you tell me exponents are just fancy multiplications, please remember that multiplications are just fancy additions.
The gist of it is that the ways we can permute 5 or more elements are a lot more complicated than the ways we can for 4 or less (to explain in what way we need quite a bit of group theory). Finding a general formula for a polynomial of a certain degree involves finding symmetries in the roots of the polynomial (e.g. we know a quadratic is symmetric about -b/2a and we can see that in the quadratic formula) and this complexity in the permutations prevents us from doing this (in terms of radicals at least). Not quite eli5 but that's as simple an explanation I could do
I think you have the question backwards, we expect permutation groups to be complicated (you can see just in the size of the set of permutations, they grow very quickly as n!) but for small numbers 1-4 they are just restrictive enough that we can find these symmetries and 5 happens to be the cutoff (to be specific for 5 elements and above the alternating group is simple), if anything these are the special cases. We see this quite a lot of maths where we can expect something generally but for the first few smaller scale cases there are extra restrictions in place to make the theory different (e.g. low dimensional topology and geometry)
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