What reduced quadratic equation has solutions \sqrt{2} and \sqrt{3}?

Step 1
A polynomial with coefficients in ${\displaystyle \mathbb{Q}}$ that has ${\displaystyle \sqrt{2}}$ as a root must also have ${\displaystyle -\sqrt{2}}$ as a root; likewise, if it has ${\displaystyle \sqrt{3}}$ as a root, then it must also have ${\displaystyle -\sqrt{3}}$ as a root.
So any polynomial with coefficients in ${\displaystyle \mathbb{Q}}$ that has both ${\displaystyle \sqrt{2}}$ and ${\displaystyle \sqrt{3}}$ as roots must be of degree at least 4 or equal to 0, and must be a multiple of ${\displaystyle (x^2-2)(x^2-3)}$.
So there is no quadratic polynomial with coefficients in ${\displaystyle \mathbb{Q}}$ that will do it.
Step 2
To see this, suppose that p(x) is a polynomial with coefficients in ${\displaystyle \mathbb{Q}}$ that has ${\displaystyle \sqrt{2}}$ as a root. Dividing with remainder by ${\displaystyle x^2-2}$ we get
${\displaystyle p\left(x\right)=(x^2-2)q\left(x\right)+r\left(x\right)}$,
where q(x) and r(x) have rational coefficients, and ${\displaystyle r\left(x\right)=0}$ or else r(x) is of degree 0 or 1.
Plugging in ${\displaystyle \sqrt{-2}}$, we get
${\displaystyle 0=0q\left(\sqrt{2}\right)+r\left(\sqrt{2}\right)=r\left(\sqrt{2}\right)}$
But there are no polynomials of degree 0 or 1 with coefficients in ${\displaystyle \mathbb{Q}}$ that are 0 at ${\displaystyle \sqrt{2}}$ (that would yield that ${\displaystyle \sqrt{2}}$ is rational), so ${\displaystyle r\left(x\right)=0}$.
Therefore, ${\displaystyle r\left(x\right)=0}$, so p(x) is a multiple of ${\displaystyle x^2-2}$; and since ${\displaystyle x^2-2}$ is 0 at ${\displaystyle -\sqrt{2}}$, then so is p(x). The same argument shows that it is also a multiple of ${\displaystyle x^2-3}$, and so p(x) will also have ${\displaystyle -\sqrt{3}}$ a a root.
Step 3
Since ${\displaystyle x^2-3}$ and ${\displaystyle x^2-2}$ are relatively ' in ${\displaystyle \mathbb{Q}\left[x\right]}$, any polynomial that is a multiple of both is a multiple of their product. The smallest degree polynomial that doe is
${\displaystyle (x^2-2)(x^2-3)=x^4-5x^2+6}$
If you allow other coefficients, then the only quadratics that work are the form
${\displaystyle c(x-\sqrt{2})(x-\sqrt{3})}$
${\displaystyle =c{x}^2-x(\sqrt{2}+\sqrt{3})x+\sqrt{6}}$
and its multiples.