Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots?

Question

Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots and has only integral parameters?

Since it has two different roots the it must be of degree at least $2$.

If it is of degree $2$, then it is of the form $(x-\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})$, but this type of polynomial has parameters which are not integers.

If it is of degree $4$ then we have the polynomial $(x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})(x-\frac{\sqrt{2}}{3})(x+\frac{\sqrt{2}}{3})$.

It remains to examine the case where it is of degree $3$. This is where I got stuck. I don't know how to examine that case. Could you please explain to me how to solve this question?

Answer 1

The polynomial with integer coefficients with least degree that has $\frac{\sqrt{3}}{2}$ as a root is $4 x^2 - 3$.

Every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ as a root is a rational polynomial multiple of $4 x^2 - 3$.

Likewise, for $\frac{\sqrt{2}}{3}$, the polynomial is $9 x^2 - 2$.

Since these two polynomials are coprime, every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{2}}{3}$ as roots is a multiple of $(4 x^2 - 3)(9 x^2 - 2)$.

Answer 2

Hint : Let $f$ be a polynomial with integer coefficients. Then if $a+\sqrt b$ is a root of $f$, where $a, b \in \Bbb Q, \sqrt b \notin \Bbb Q$, then $a-\sqrt b$ is also a root of $f$.