Suppose that a real quartic (= degree 4) polynomial had a positive discriminant.

Question

Suppose that a real quartic (= degree 4) polynomial had a positive discriminant. What can you say about the number of real roots?
Suppose that a real quartic degree 4 polynomial has a positive discriminant. What can you say about the number of real roots

Explanation / Answer

If there is a double root (a repeated root) of your polynomial, then the discriminant will be zero. (And, likewise, a zero discriminant means that there is a repeated root.) Now suppose your coefficients are all real numbers, and there happen to be four real roots (p, q, r, s). Then the discriminant is [(p-q)(p-r)(p-s)(q-r)(q-s)(r-s)]2 > 0. But if you have two real roots and two complex roots (p, q, r+si, r-si)? Then the discriminant is (p-q)2 [(p-r-si)(p-r+si)]2 [(q-r-si)(q-r+si)]2 (2si)2 = (p-q)2 ((p-r)2 + s2)2 ((q-r)2 + s2)2 * (-4s2) < 0. Similarly, if you have four complex roots (p+qi, p-qi, r+si, r-si), then the discriminant will be positive. So you may conclude that if the discriminant is negative, then you have two real roots and two complex roots. You have to do a little more work to distinguish between four real roots and four complex roots, but there are various ways of doing that once you know that the discriminant is positive. I'd have to work on coming up with one (probably consulting Dummit and Foote in the process), so I'll stop here for now and see if this is the kind of thing you want.

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