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is there a method to find all the solutions to the following set of irrational equations,

$\sqrt{x}+y=3$

$x+\sqrt{y}=5$

NOTE: $(4-1)=(2-1)(2+1)=3$ and $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$

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    @Graphth: ?? And so what?2012-08-27

4 Answers 4

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$(3-y)^2+\sqrt{y}=5$

$y^2-6y+\sqrt{y}+4=0$

$(\sqrt{y}-1)(y\sqrt{y}+y-5\sqrt{y}-4)=0$

i.e. $\sqrt{y}=1$, and $x=4$ is a solution

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    You should probably edit your question and add this there to show how everything relates and show your effort. For future reference, diophantine equations involve finding rational or integer solutions to equations. I'll change my vote on the question and retag it to something more appropriate.2012-08-21
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...is there a method to find all the solutions[?]

Yes, there is: draw the graphs of the functions $x\mapsto3-\sqrt{x}$ and $x\mapsto(5-x)^2$ on $[0,5]$, these intersect at $(4,1)$ and there only hence the unique solution of your system is $(x,y)=(4,1)$.

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For the question in the title, you can move the $\sqrt y$ to the other side, square, and get a quartic which yields to the rational root theorem.

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    Which says nothing about possible other roots.2012-08-22
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If you make the change of variables $\begin{equation*} u=\sqrt{x}\geq 0,\qquad v=\sqrt{y}\geq 0, \end{equation*}$

then you need to solve $\begin{equation*} \left\{ \begin{array}{c} u+v^{2}=3 \\ u^{2}+v=5 \end{array} \right. \end{equation*}$

or $\begin{eqnarray*} &&\left\{ \begin{array}{c} u=3-v^{2} \\ v^{4}-6v^{2}+v+4=0 \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} u=3-v^{2} \\ \left( v-1\right) \left( v^{3}+v^{2}-5v-4\right) =0. \end{array} \right. \end{eqnarray*}$

So $u=2,v=1$ is a solution, i.e. $x=4,y=1$ in the original variables. And since $v^{3}+v^{2}-5v-4=0$ has 3 real solutions, one positive $v_{1}\approx 2.164\,2\gt \sqrt{3}$ and two negative $v_{2}\approx -2.391\,4,v_{3}\approx -0.772\,87$, there are no other solutions because $u=3-v_{1}^{2}<0$.