Descartes' rule of signs tells us that x^4 - x^3 -1 =0 has exactly 1 positive

root, and putting x=-y to get y^4+y^3-1=0, allows us to again apply the

rule of signs to tell us that the original equation has exactly 1 negative

root.

So x^4 - x^3 -1 =0 has two real roots.

(Descartes rule of signs may be found here. Note in this case it tells

us exactly how many roots there are)

RonL