Afriend tells you that he has a cubic equation with exactly three complex roots. determine which explanation best explains why this is impossible. a) cubic equations must have all real roots and no complex solutions. b) there must be only two real solutions to the equation. c) complex solutions must appear in conjugate pairs; having an odd number of them is impossible. d) cubic equations cannot have any complex solutions

Complex solutions, namely roots with a √(-1) or "i" in it, never come all by their lonesome, because an EVEN root like the square root, can have two roots that will yield the same radicand.

a good example for that will be √(4), well, (2)(2) is 4, so 2 is a root, but (-2)(-2) is also 4, therefore -2 is also a root, so you'd always get a pair of valid roots from an even root, like 2 or 4 or 6 and so on.

therefore, complex solutions or roots are never by their lonesome, their sister the conjugate is always with them, so if there's a root a + bi, her sister a - bi is also coming along too.

if complex solutions come in pairs, well, clearly a cubic equation can't yield 3 only.

If i could just have the statements, i'd be able to