Hi,
My question is how do you solve analitically:
sqrt(x)+y=7 and sqrt(y)+x=11 without solving a 4th degree polynomial equation?
Thanks
Dear kolg,
$\displaystyle \sqrt{x}+y=11$------------(1)
$\displaystyle \sqrt{y}+x=7$-------------(2)
From (1); y<11 and $\displaystyle x\geq{0}$
From (2); x<7 and $\displaystyle y\geq{0}$
Therefore, $\displaystyle 0\leq{y}<11$ and $\displaystyle 0\leq{x}<7$
Also from (1) and (2) it could be seen that, $\displaystyle \sqrt{x}\in{N}$ and $\displaystyle \sqrt{y}\in{N}$
Therefore x=4 is the only solution for x.
When, $\displaystyle x=4\Rightarrow{y=9}$
Therefore x=4, y=9 is the only solution for this system of equations.
Dear Klog,
If $\displaystyle \sqrt{x}$ is rational but not an integer,
$\displaystyle \sqrt{x}=\frac{p}{q}$ where $\displaystyle q\neq{1}$
$\displaystyle x=\frac{p^2}{q^2}$
Therefore, $\displaystyle \sqrt{y}=7-\frac{p^2}{q^2}=(\sqrt{7}-\frac{p}{q})(\sqrt{7}-\frac{p}{q})$
Hence $\displaystyle \sqrt{y}$ is irrational.
Then $\displaystyle x+\sqrt{y}$ is irrational. (Contradiction)
Therefore $\displaystyle \sqrt{x}$ is irrational or an integer.(according to our assumption.)
But if $\displaystyle \sqrt{x}$ is irrational $\displaystyle y+\sqrt{x}$ is irrational. (Contradiction)
So $\displaystyle \sqrt{x}$ is an integer.
Same proof goes to $\displaystyle \sqrt{y}$ as well.
Hope this helps.