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Does there exist a reduced quadratic equation with solutions equal to both the $\sqrt{2}$ and $\sqrt{3}$?

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Where do the coefficients lie? –  awllower Mar 2 '11 at 12:29
    
Is it acceptable to say that since $sqrt{2}$ and $sqrt{3}$ are not conjugate, it is impossible to find it? –  awllower Mar 2 '11 at 12:30
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@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... –  Arturo Magidin Mar 2 '11 at 17:22
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3 Answers 3

up vote 7 down vote accepted

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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This expands the meaning of the comments above, and truly, if the coefficients are not properly limited, then the question leaves nothing to be done. This might be the problem of formulation, and I would like to see what OP wants to say about that. –  awllower Mar 3 '11 at 5:18
    
In any case, this is a good answer, @Arturo Magidin. –  awllower Mar 3 '11 at 5:19
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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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But then the coefficients are not rational... –  awllower Mar 2 '11 at 13:16
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@awllower: True, but there is no quadratic with rational coefficients and roots $\sqrt{2}$ and $\sqrt{3}$. There is a quartic: $(x^2-2)(x^2-3)$. Your question of where the coefficients lie is a good one. –  Ross Millikan Mar 2 '11 at 14:35
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@awllower, I don't see any mention of rational coefficients in the question, unless it's in "reduced". –  Peter Taylor Mar 2 '11 at 15:00
    
well we also have "equation have" which isnt conjugated properly... –  yoyo Mar 2 '11 at 16:56
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$x^2-(\sqrt2+\sqrt3)x+\sqrt6=0$

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