Question

solve he equation x^3-x-1=0
show your method help!!!

Answer 1

x^3-x-1
Taking x common :
x(x^2-1)-1=0
Therefore : x=+-1

Answer 2

It has no rational roots. 1 and -1 do not work.

Answer 3

you mean by x=+-1????

Answer 4

He means positive 1 and negative 1

Answer 5

not a solution right?

Answer 6

Right. Neither one of them make the sentence true.

Answer 7

There are no rational roots.

Answer 8

how u know that?

Answer 9

The possible rational roots are the factors of the constant term divided by the factors of the coefficient of the leading term. Since both numbers are 1, the only possible rational roots are 1 and negative 1 and neither of those is a root.

Answer 10

Do you know the rational roots theorem?
e Rational Roots (or Rational Zeroes) Test is a handy way of obtaining a list of useful first guesses when you are trying to find the zeroes (roots) of a polynomial. Given a polynomial with integer (that is, positive and negative "whole-number") coefficients, the possible (or potential) zeroes are found by listing the factors of the constant (last) term over the factors of the leading coefficient, thus forming a list of fractions. This listing gives you a list of potential rational (fractional) roots to test -- hence the name of the Test.

Answer 11

ok! any idea how to do it?

Answer 12

I would test values to see where it crosses the x axis and then drill down.

Answer 13

If x = 0, y is negative
if x = 1, y is negative
if x = 2, y is positive
So there is a root between 1 and 2
if x = 1.5, y is positive so there is a root between 1 and 1.5

Answer 14

it did actually for x between 1 and 1.5

Answer 15

You could graph it with a graphing calculator and find the root also.

Answer 16

with a graphing calculator they just give the solution approximately
i do more care about the exact form of the solution and the method
i need it to fin padovan's general term

Answer 17

You could say: x^3-x=1
3st = -1 s=-1/3t
s^3-t^3=1

Answer 18

sorry i didnt understand what u wrote

Answer 19

It is a method of changing the equation into a quadratic form with t^6 and t^3 so that it can be solved using the quadratic formula.