Thread: number of real roots?

The number of real solutions to is?

-- ok, so i figured that cosec^2(x) = 1 + cot^2(x), so that brings it to
if i am right?

so how many solutions are there then? 4?

EDIT: using a graphics calc, i graphed x^4 - x^3 -1 =0 and checked how many x intercepts there were, and so there were 2 real solutions.
But is there a way of doing this without the calculator?

Originally Posted by scorpion007

The number of real solutions to is?

-- ok, so i figured that cosec^2(x) = 1 + cot^2(x), so that brings it to
if i am right?

so how many solutions are there then? 4?

EDIT: using a graphics calc, i graphed x^4 - x^3 -1 =0 and checked how many x intercepts there were, and so there were 2 real solutions.
But is there a way of doing this without the calculator?

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