Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
@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
1  
@Mike And, yet, the question has 3 upvotes and is favorited by 1 user. –  Graphth Aug 21 '12 at 16:40
show 1 more comment

4 Answers 4

up vote 1 down vote accepted

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 $u=3-v_{1}^{2}<0$.

share|improve this answer
add comment

...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 $(x,y)=(4,1)$.

share|improve this answer
add comment

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.

share|improve this answer
    
Which says nothing about possible other roots. –  Did Aug 22 '12 at 11:28
add comment

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

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

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

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

share|improve this answer
1  
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. –  Mike Aug 21 '12 at 15:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.