Does there exist a reduced quadratic equation with solutions equal to both the $\sqrt{2}$ and $\sqrt{3}$?
What reduced quadratic equation has solutions $\sqrt{2}$ and $\sqrt{3}$
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1@awllower: it is impossible to find a quadratic with coefficients in $\mathbb{Q}$ that will do it; but it is not impossible to find one *somewhere*... – 2011-03-02
3 Answers
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 $(x^2-2)(x^2-3)$. So there is no quadratic polynomial with coefficients in $\mathbb{Q}$ that will do it.
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 $p(x) = (x^2-2)q(x) + r(x)$, where $q(x)$ and $r(x)$ have rational coefficients, and $r(x)=0$ or else $r(x)$ is of degree 0 or 1. Plugging in $\sqrt{-2}$, we get $0 = 0q(\sqrt{2}) + r(\sqrt{2}) = r(\sqrt{2})$. 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 $r(x)=0$. Therefore, $r(x) = 0$, 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 $-\sqrt{3}$ as a root.
Since $x^2-3$ and $x^2-2$ are relatively prime in $\mathbb{Q}[x]$, any polynomial that is a multiple of both is a multiple of their product. The smallest degree polynomial that does is $(x^2-2)(x^2-3) = x^4 - 5x^2 + 6$.
If you allow other coefficients, then the only quadratics that work are of the form $c(x-\sqrt{2})(x-\sqrt{3}) = cx^2 - c(\sqrt{2}+\sqrt{3})x + \sqrt{6}$ and its multiples.
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0In any case, this is a good answer, @Arturo Magidin. – 2011-03-03
A quadratic equation with roots $a$ and $b$ is of the form $c(x-a)(x-b)$. I assume that "reduced" here means that you take $c = 1$.
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0well we also have "equatio$n$ have" which isnt conjugate$d$ properly... – 2011-03-02
$x^2-(\sqrt2+\sqrt3)x+\sqrt6=0$