First, let's simplify the given equation:
x^3 - 2x + 5 = x^3 - x^2 + 9 x^2 - 2x - 4 = 0
We can use the quadratic formula to solve for x:
x = (2 ± √(2^2 - 4(1)(-4))) / 2 x = 1 ± √5
Since r is a root of x^3 - 2x + 5, we know that r^3 - 2r + 5 = 0.
Now, let's assume that r is irrational. Then we know that r can be expressed as a non-repeating, non-terminating decimal.
However, if we cube both sides of the equation r^3 - 2r + 5 = 0, we get:
r^3 = 2r - 5
Substituting in the decimal expression for r, we get:
(decimal expression for r)^3 = 2(decimal expression for r) - 5
This means that the cube of the decimal expression for r can be expressed as a finite sum of rational numbers (2 times the decimal expression for r, minus 5).
But this is a contradiction, since the cube of any non-terminating, non-repeating decimal is also a non-terminating, non-repeating decimal (and therefore cannot be expressed as a finite sum of rational numbers).
Therefore, r cannot be irrational.
Similarly, we can show that neither r^2 nor r^3 is irrational. If r^2 were irrational, then r^2 could be expressed as a non-repeating, non-terminating decimal. But then r^3 = (r^2)r would also be a non-terminating, non-repeating decimal, which contradicts our earlier conclusion.
Finally, if r^3 were irrational, then r^3 could be expressed as a non-repeating, non-terminating decimal. But then r = (r^3)/(2 - 5) would also be a non-terminating, non-repeating decimal, which again contradicts our earlier conclusion.
Therefore, none of r, r^2, or r^3 can be irrational.
