# is there a method to find all the solutions to, $y^2-6y+\sqrt{y}+4=0$

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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The title does not match the question and neither of which match the tag. The note seems completely irrelevant and pointless unless you're trying to lead people in a specific direction. –  Mike Aug 21 '12 at 14:28
Maybe then just saying "Yes, there is" is the right answer, since it applies to both questions. –  Zarrax Aug 21 '12 at 14:32
the note indicates that $\sqrt{x}=2,y=1$ is one of the solution(s). –  Rajesh K Singh Aug 21 '12 at 14:34
@RajeshKSingh: If the equation in the title is what you got by manipulating one equation and then substituting into the other equation, you should right it explicitly as a solution attempt. –  user3533 Aug 21 '12 at 14:35
@Mike And, yet, the question has 3 upvotes and is favorited by 1 user. –  Graphth Aug 21 '12 at 16:40
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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 $u=3-v_{1}^{2}<0$.

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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. –  Did Aug 22 '12 at 11:28
$(3-y)^2+\sqrt{y}=5$
$y^2-6y+\sqrt{y}+4=0$
$(\sqrt{y}-1)(y\sqrt{y}+y-5\sqrt{y}-4)=0$
i.e. $\sqrt{y}=1$, and $x=4$ is a solution