Talk:Quadratic equation/Advanced

"Fortunately, there exists a simple closed formula ..." Well, yes, but there exists a complicated closed formula for cubic equations, but that's not all that fortunate as in the general case it requires finding the cube root of a complex number, which is not a straightforward operation like the square root. In fact it seems to be much the same as solving a cubic equation. I think the closed form isn't the relevant point. Peter Jackson 18:04, 4 December 2008 (UTC)

The words "there exists a closed formula" already were there, and I added the words "Fortunately" and "simple". I put the word fortunate, because once it was known that 5th and higher degree equations were not generally solvable by radicals, it does seem lucky that the most common cases can be solved such, and in fact there is one formula that works for all equations. I put simple because it is much simpler than the cubic (although not much simpler than the reduced cubic) formula, and much much much much simpler than the quartic formula.

I'm not sure what your objection is -- is it "fortunate", "simple", "closed formula", or the whole bit? If "closed formula" isn't the relative point, what is? When you say "it seems to be much the same as solving a cubic equation", what is "it"?Barry R. Smith 18:18, 4 December 2008 (UTC)

"It" is taking the cube root of a complex number. You can solve the general quadratic with real coefficients by "simple" arithmetic operations, including taking square roots of real numbers. There is no such solution for the general cubic with real coefficients. Peter Jackson 18:27, 4 December 2008 (UTC)