Prove that $(x-1-\sqrt{2-i \sqrt3})(x-1+\sqrt{2+i\sqrt{3}})$ is not in $\mathbb{R}[x]$

Question

Prove that $(x-1-\sqrt{2-i \sqrt3})(x-1+\sqrt{2+i\sqrt{3}})$ is not in $\mathbb{R}[x]$

Solution: $$ p(x) = ((x-1)-\sqrt{2-i \sqrt3})((x-1)+\sqrt{2+i\sqrt{3}})=(x-1)^{2}\color{red}{-}(\sqrt{2+i\sqrt{3}})^{2}= (x^{2}+2x+1)\color{red}{-}(2+i\sqrt{3}) $$ so, the polynomial $p(x)$ is not in $\mathbb{R}[x]$.

Please check that this solution is correct. I tried prepare other solution, but am not idea.

Answer 1

There is no need to expand all the terms—doing so is algebraically messy and ultimately led to your error.

A monic quadratic polynomial with roots $\alpha,\beta \in \mathbb{C}$ is in $\mathbb{R}[x]$ if and only if both $\alpha + \beta$ and $\alpha\beta$ are real, since we have $(x-\alpha)(x-\beta) = x^2-(\alpha+\beta)x + \alpha\beta$.

Apply this with $\alpha = 1+\sqrt{2-i\sqrt{3}}$ and $\beta = 1-\sqrt{2+i\sqrt{3}}$. Is $\alpha+\beta$ real?

Answer 2

$(a-b)(a+b)=a^2-b^2$. You wrote instead $a^2b^2$. There must be a missing $-$ sign in between. Otherwise it is correct.

You could also prove it without computing. Real polynomials have roots that come in pairs that are complex conjugate of each other. The complex conjugate of $1+\sqrt{2-i\sqrt{3}}$ is $1+\sqrt{2+i\sqrt{3}}$, which is different from $1-\sqrt{2+i\sqrt{3}}$.

It is current agreement that the symbol $\sqrt{x}$ denotes the principal value of the square root, not two values. The symbol $x^{1/2}$ is reserved for that. So, your question is properly formulated.