How Do You Solve the Equation x^3 - x - 1 = 0?

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How to solve this? Please help!
 
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By the "rational root theorem", which says that if m/n (m and n integers) is a rational root of [itex]a_nx^n+ a_{n-1}x^{n-1}+\cdot\cdot\cdot+ a_1x+ a_0[/itex] then n must divide the "leading coefficient", [itex]a_n[/itex], and m must divide the "constant term", a_0. That tells us that the only possible rational roots for this equation are the intgers 1 and -1. Unfortunately, trying them [itex]1^3- 1- 1= -1[/itex] and [itex](-1)^3- (-1)-1= -1[/itex]. Since neither of those is equal neither 1 nor -1 is a root and so this equation has no rational roots.

But if x= 2, [itex]2^3- 2- 1= 5[/itex] so there is clearly a root between 1 and 2. The only thing I could suggest is Cardano's "cubic formula": http://www.sosmath.com/algebra/factor/fac11/fac11.html.
 
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i have a similar problem

x^2 -x -20 < 0

this is what i have so far:

x^2 - x < 20

x^2 - x < 20
_______x to eliminate the power

x - 1 < 20
_______x


i'm pretty much stuck there, any help?
p.s. underscore is just to put the X where i want it
 


What exactly are you to do with that inequality? Find the range of values for x?



x - 1 < 20
_______x

I think at one point you divided by x to get this, but you can't just divide by x in an inequality, as the sign of x can change the inequality.

But if you want to get [itex]x^2-x-20[/itex] to have x to a single power then try completing the square.
 


Epic Jeff said:
i have a similar problem

x^2 -x -20 < 0

this is what i have so far:

x^2 - x < 20

x^2 - x < 20
_______x to eliminate the power

x - 1 < 20
_______x


i'm pretty much stuck there, any help?
p.s. underscore is just to put the X where i want it
Please do not "hijack" other people's threads for a new problem. And, I can see no similarity, except that they both involve polynomials.

I recommend you factor [itex]x^2- x- 20[/itex] which can be done relatively easily. Use the fact that the product of two numbers is negative if and only if one is positive and the other negative.
 


My bad, thanks for the help though.