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 14, 2023

#1**-1 **

Let's assume that r is an irrational number. Then, since r is a root of the equation x^3 - 2x + 5 = x^3 - x^2 + 9, we have:

r^3 - 2r + 5 = r^3 - r^2 + 9

Simplifying this equation, we get:

r^2 = 7

This implies that r is a root of the quadratic equation:

x^2 - 7 = 0

Therefore, r must be a rational number, which contradicts our assumption that r is irrational. Hence, r cannot be irrational.

Now, let's assume that r^2 is irrational. Then, since r is a root of the equation x^3 - 2x + 5 = x^3 - x^2 + 9, we have:

r^3 - 2r + 5 = r^3 - r^2 + 9

Substituting r^2 = 7, we get:

r^3 - 2r + 5 = r^3 - 7 + 9

Simplifying this equation, we get:

r = 1

This implies that r is a rational number, which contradicts our assumption that r^2 is irrational. Hence, r^2 cannot be irrational.

Finally, if r^3 were irrational, then we would have:

r^3 = 2r - 5 + r^2 - 9

Substituting r^2 = 7, we get:

r^3 = 2r - 7

This implies that r is a root of the quadratic equation:

x^2 - 2x + 7 = 0

Using the quadratic formula, we get:

x = 1 ± 2i

Therefore, r cannot be a real number, which contradicts our assumption that r is a root of a real polynomial. Hence, r^3 cannot be irrational.

In conclusion, we have shown that none of r, r^2, or r^3 can be irrational if r is a root of the equation x^3 - 2x + 5 = x^3 - x^2 + 9.

Anthonyward Mar 14, 2023