William Collins

2022-01-13

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

aquariump9

Expert

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

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