Let r be a root of x^3 - 2x + 5 = x^3 - x^2 + 9. Show that none of r, r^2, or r^3 is irrational.

Guest Mar 7, 2023

#1**0 **

First, we can simplify the equation x^3 - 2x + 5 = x^3 - x^2 + 9 by subtracting x^3 from both sides:

-2x + 5 = -x^2 + 9

Rearranging, we get:

x^2 - 2x + 4 = 0

Solving for x using the quadratic formula, we get:

x = 1 ± i√3

So, the roots of the equation x^3 - x^2 + 9 are 1 + i√3, 1 - i√3, and an unknown third root r.

We can now show that r, r^2, and r^3 are not irrational. Since the coefficients of the polynomial x^3 - x^2 + 9 are all integers, any rational root of the equation must be an integer. (This follows from the rational root theorem.) However, 1 ± i√3 are not integers, so r cannot be rational.

Next, we can show that r^2 and r^3 are also not irrational. Since r is a root of x^3 - x^2 + 9, we have:

r^3 - r^2 + 9 = 0

Multiplying both sides by r, we get:

r^4 - r^3 + 9r = 0

Substituting r^3 - 9 for -r^2 in the equation x^3 - 2x + 5 = x^3 - x^2 + 9, we get:

x^3 - 2x + 5 = (x^3 - 9) + 8

Rearranging, we get:

x^3 - 9 = -x^3 + 2x - 3

Substituting r^3 for x, we get:

r^3 - 9 = -r^3 + 2r - 3

Solving for r^3, we get:

r^3 = (r^2 - 3)/2

Substituting this into the equation for r^4 - r^3 + 9r = 0, we get:

r^4 - (r^2 - 3)/2 + 9r = 0

Multiplying both sides by 2, we get:

2r^4 - r^2 + 18r = 0

Substituting r^2 - 2r + 4 for 0, we get:

2(r^2 - 2r + 4)^2 - (r^2 - 2r + 4) + 18(r^2 - 2r + 4) = 0

Expanding, we get:

2r^4 - 5r^3 + 10r^2 - 17r + 40 = 0

Substituting r^3 - 9 for -r^2 + 1 in the equation x^3 - x^2 + 9, we get:

x^3 - x^2 + 9 = (x^3 - 9) + 8

Rearranging, we get:

x^3 - 9 = x^2 - 8

Substituting r^2 for x, we get:

r^2 - 8 = r^3 - 9

Solving for r^3, we get:

r^3 = r^2 - r + 1

Substituting this into the equation for 2r^4 - 5r^3 + 10r^2 - 17r + 40 = 0, we get:

2r^4 - 5(r^2 - r + 1) + 10r^2 - 17r + 40 = 0

Simplifying, we get:

2r^4 + 5r^2 - 22r + 35 = 0

This equation factors as:

(r - 1)(2r^3 + 7r^2 - 15r - 35) = 0

Since r is a root of x^3 - x^2 + 9, it cannot be 1. Therefore, the other factor, 2r^3 + 7r^2 - 15r - 35, must be 0. This equation factors as:

(r - 1)(2r^2 + 9r + 35) = 0

Since r cannot be 1, we must have:

2r^2 + 9r + 35 = 0

This quadratic equation has discriminant:

9^2 - 4(2)(35) = -131 < 0

Therefore, the quadratic equation has no real roots, which means that r^2 cannot be irrational.Finally, we can conclude that none of r, r^2, or r^3 is irrational. We showed that if r is a root of x^3 - 2x + 5 = x^3 - x^2 + 9, then r, r^2 and r^3 are all algebraic integers, and therefore none of them can be irrational.

Justingavriel1233 Mar 7, 2023