# Irrational polynomial equation

1. Oct 19, 2007

### ciel

what's the polynomial equation which sqrt2 + sqrt3 satisfies ?

2. Oct 19, 2007

### ZioX

3. Oct 20, 2007

### Diffy

There is not one unique polynomial that (sprt2 + sqrt3) satisfies. Perhaps you should elaborate a bit as to what terms, degree, etc. that you require in your polynomial.

4. Oct 20, 2007

### Hurkyl

Staff Emeritus
And keep in mind that a polynomial is nothing more than a linear combination of monomials...

5. Oct 21, 2007

### Gib Z

What about, $$x- \pi$$ ? or $$x^3 - 31$$ =] ?

6. Oct 21, 2007

### AlphaNumeric2

If you want to get rid of the roots, then it'd be something like

$$x = \sqrt{2}+\sqrt{3}$$
$$x^{2} = 2+3+2\sqrt{6}$$
$$x^{2}-5 = 2\sqrt{6}$$
$$(x^{2}-5)^{2} = 24$$

Rearrange and clean up a bit.

7. Oct 21, 2007

### ciel

yeah, i seems alright that way. well, i has to be polynomial, but i couldn't just go upto that part, so i was confused. thnx anyway :)

8. Oct 24, 2007

### EES

9. Oct 24, 2007

### robert Ihnot

Hurkyl: And keep in mind that a polynomial is nothing more than a linear combination of monomials...

From the standpoint of Galois Theory, we can build that form from the conjugates of the two order two equations, which produces an order 4 equation:

The four products of the form: $$\prod (X-(\pm\sqrt2\pm\sqrt3)$$ = X^4-10X^2+1.

Last edited: Oct 24, 2007
10. Oct 25, 2007

### HallsofIvy

It would have been nice if you had posted the actual question. I suspect it was NOT "Find a polynomial equation that $\sqrt{2}+ \sqrt{3}$ satisfies". I suspect rather that it was something like "Find a polynomial equation, with integer coefficients, that $\sqrt{2}+ \sqrt{3}$ satisfies".

11. Oct 25, 2007

### Gib Z

I'm just joking with you man, try it on yours calculator, $\sqrt{2}+\sqrt{3}$ is a valid approximation of pi to several digits. Similar thing for the other equation.