Alright I was helping my friend with some high school math and she had the following question:
if 2 + sqrt(3) is a polynomial root, name another root of the polynomial, and explain how you know it must also be a root

Question

Alright I was helping my friend with some high school math and she had the following question:
if 2 + sqrt(3) is a polynomial root, name another root of the polynomial, and explain how you know it must also be a root

anonymous · Answer

I know the answer should be:
$$2 - \sqrt3$$
but here is the thing....
the question never gives a polynomial. It just gives a condition that the polynomial must statisfy. Well doesn't
$$(x -2 - \sqrt3)^2$$
satisfy that ciondition? Yet the only root of this polynomial is
$$2 + \sqrt3$$

anonymous · Answer

so basically I managed to confuse myself on something basic...I know its something I am forgetting but I don't know what.

anonymous · Answer

the original question isn't 2+i*sqrt(3) and all coef are real?

anonymous · Answer

http://imgur.com/A4D2BYl

hero · Answer

The basis for the question itself comes from a property of radical conjugate roots. If a polynomial $P(x)$ with rational coefficients has $a + \sqrt{b}$ as a root, where a, b are rational and $\sqrt{b}$ is irrational, then $a − \sqrt{b}$ is also a root. http://en.wikipedia.org/wiki/Properties_of_polynomial_roots#Radical_conjugate_roots

anonymous · Answer

I found that also which is why I know what the answer should be but I still can't see what is wrong with my possible polynomial

anonymous · Answer

There has to be something wrong with my proposition because it seems to defy the property of radical conjugate roots.

anonymous · Answer

@Hero but he dont mention taht the coef are rational

anonymous · Answer

lol at least I'm not the only one this is a problem for lol

hero · Answer

The product of $(x - 2 - \sqrt{3})^2$ produces co-efficients that are not rational.

anonymous · Answer

...so a polynomial has to have rational coefficients?

hero · Answer

Technically, since the question itself doesn't mention anything about rational roots, your response could also be accepted, but you should present your argument in the following manner:

Assume $x = 2 + \sqrt{3}$
Then $x - 2 - \sqrt{3} = 0$
and $(x - 2 - \sqrt{3})^2 = 0$
yet the graph of $y = (x - 2 - \sqrt{3})^2 $ produces only one root.

hero · Answer

*rational coefficients not rational roots

anonymous · Answer

....but in the wiki you linked it says nothing about the coefficients being rational

anonymous · Answer

it says the coefficents are members of the field of complex numbers which, excuse me if I'm wrong, the coefficents of my polynomial are

anonymous · Answer

@Hero please answer, I'm still a little confused

hero · Answer

I'll link you to it once more. You have to align your eyes to the correct area on the page. Look for the words "rational coefficients" while scanning the page. http://en.wikipedia.org/wiki/Properties_of_polynomial_roots#Radical_conjugate_roots

anonymous · Answer

...oops lol I'll blame the fact that the page said this on the first line:
where the ai belong to some field, which, in this article, is always the field \mathbb C of the complex numbers

anonymous · Answer

I get it now thanks :P

hero · Answer

I think you were looking at the wrong area. I linked you a specific area on the page.

hero · Answer

The part that is entitled "Radical Conjugate Roots"

anonymous · Answer

Yup I was. Thanks again. Just a poorly written question on her instructor's part (not surprising with this guy). I just didn't see that this time.