# How many complex roots does 6x^5+x^4-3=0 have?

Sep 13, 2014

Solve: 6x^5+x^4−3=0
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