How many complex roots does #6x^5+x^4-3=0# have?
1 Answer
Solve:
How many real and/or complex roots does it have?
Here is the graph:
It appears that the function crosses the x-axis exactly once. That would mean that the equation has 1 real and 4 imaginary (complex) solutions.
I consulted WolframAlpha to be sure, but then I also decided to try out Descartes' Rule of Signs! (an oldie but goodie)
f(x) goes + + - in terms of its "signs" on the terms
This pattern shows just one sign change from + to -, so this predicts that there will be exactly ONE positive, real zero.
f(-x) would be - + - showing 2 sign changes. This predicts either 2 or 0 negative real zeros. Since the graph reveals 0 negative real zeros, then we must have 1 positive real zero, 0 negative real zeros, and 4 complex zeros. Phew!
Here is a link to Purple Math for a little more explanation of this idea...Purple Math