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

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Homework Help Overview

The discussion centers around solving the equation x3 - x - 1 = 0, which falls under the topic of polynomial equations. Participants explore various methods to identify roots of the cubic equation and discuss related polynomial inequalities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the rational root theorem and the potential for dividing the polynomial by linear binomials. Some suggest testing specific values as potential roots, while others note the absence of rational roots based on their calculations. There is also a mention of Cardano's cubic formula as a possible approach.

Discussion Status

The discussion is ongoing, with various methods being proposed and explored. Some participants are seeking clarification on the steps involved in solving the cubic equation, while others are addressing a related inequality problem, prompting questions about the correct approach to inequalities.

Contextual Notes

One participant raises a separate inequality problem, which leads to a discussion about the implications of dividing by a variable in inequalities. This highlights the complexity of handling different types of polynomial expressions and their solutions.

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How to solve this? Please help!
 
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If you can divide the polynomial expression by a linear binomial, then you can determine what value of x of each binomial will make the binomial equal to zero. Try first divisors x+1, and x-1.
 


By the "rational root theorem", which says that if m/n (m and n integers) is a rational root of a_nx^n+ a_{n-1}x^{n-1}+\cdot\cdot\cdot+ a_1x+ a_0 then n must divide the "leading coefficient", a_n, 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 1^3- 1- 1= -1 and (-1)^3- (-1)-1= -1. 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, 2^3- 2- 1= 5 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 x^2-x-20 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 x^2- x- 20 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.
 

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