Find f(x) such that one root is x = sqrt 2 + cube-root 2

THE COEFFICIENTS MUST BE INTEGERS
use the conjugate root theorem a+b*sqrt(c) (in this case I think a is cube-root 2, and b*sqrt(c) is sqrt2
PLEASE DON'T GIVE ME A B.S. ANSWER THX

Let x = 2^(1/2) + 2^(1/3)
==> x - 2^(1/2) = 2^(1/3).

Cube both sides:
x^3 - 3x^2 * 2^(1/2) + 3x * 2 - 2^(3/2) = 2
==> x^3 - 3x^2 * 2^(1/2) + 6x - 2 * 2^(1/2) = 2
==> x^3 + 6x - 2 = (3x^2 + 2) * 2^(1/2).

Square both sides:
x^6 + 12x^4 - 4x^3 + 36x^2 - 24x + 4 = (9x^4 + 12x^2 + 4) * 2
==> x^6 - 6x^4 - 4x^3 + 12x^2 - 24x - 4 = 0.

(Double checked with Wolfram Alpha.)

I hope this helps!

Not too long..., once I was finally able to sit down (from teaching classes all day) and finally answer it. Admittedly, I originally made an algebraic mistake toward the end, which I caught by double checking the equation's solution on Wolfram Alpha.

Good luck!

Report Abuse