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Equations such as $\sqrt{x+1}+\sqrt{x+2}=x+3$ are easily solvable by squaring both sides. But if we increase an extra square root, like if trying to solve $\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}=x+4$ we encounter a problem. Squaring both sides doesn't work now, since the L.H.S. would still end up with three terms involving square roots, rather than only one as for the previous equation.

That is, unless I'm missing something. Is there a way of solving equations such as the following analytically?

$\sqrt{P_1(x)}+\sqrt{P_2(x)}+\cdots+\sqrt{P_n(x)}=Q(x)$ where $n>2$ and $Q(x), P_1(x), \ldots, P_n(x)$ are all polynomials.

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You can solve your second example by taking the term $\sqrt{x+3}$ to the RHS and then squaring --- only two radicals are left then. The same trick doesn't work for $n$ terms though. –  Federico Poloni Nov 8 '12 at 7:28
    
You're right! Good insight. –  Alexander Farrugia Nov 8 '12 at 7:33
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up vote 10 down vote accepted

Let us set $z_j=\sqrt{P_j(x)}$, so that the equation becomes a system of algebraic equations: $$z_1+\cdots+z_n=Q(x),\qquad P_j(x)=z_j^2\qquad\forall j=1,\ldots,n.$$ Using the resultant, you may eliminate the $z_j$'s. At the end, you obtain a single polynomial equation $R(x)=0$. Now, the difficulty is that $R$ has degree $\ge5$ (except in your solvable example, where it has degree $4$), and unless your data is really exceptional, this equation is not solvable, in the sense of Galois's theory. For instance, if $n=3$, the elimination of $z_3$ gives $(z_1+z_2-Q(x))^2=P_3(x)$. Then the elimination of $z_2$ yields an equation of degree $\ge4$ and the last elimination, of $z_1$ gives an $R$ of degree $\ge8$.

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Thank you very much for this answer. –  Alexander Farrugia Nov 8 '12 at 7:53
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Everything depends on the precise meaning of "solving analytically". One can use other functions besides radicals, for example, elliptic or theta-functions. –  Alexandre Eremenko Nov 8 '12 at 12:55
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