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- Feb 14, 2012
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Find the real solution(s) to the equation \(\displaystyle \frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}\).
Hey anemone!Find the real solution(s) to the equation \(\displaystyle \frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}\).
Hi I like Serena,Hey anemone!
Nice little puzzle.
if $\sqrt{x}=3$ \(\displaystyle \frac{36}{3}+9(3)+\frac{9}{\sqrt{y}}+ \sqrt{y}=42\) \(\displaystyle \frac{9}{\sqrt{y}}+ \sqrt{y}=3\) \(\displaystyle 9=\sqrt{y}(3-\sqrt{y})\) and this equation has no real solution for $y$. | if $\sqrt{x}=4$ \(\displaystyle \frac{36}{4}+9(4)+\frac{9}{\sqrt{y}}+ \sqrt{y}=42\) \(\displaystyle \frac{9}{\sqrt{y}}+ \sqrt{y}=-3\) and this equation has no real solution for $y$. |
I'm glad to see you enjoy math so much.Hi I like Serena,
I want to thank you for participating and thanks for the compliment to this problem as well..., I'm so flattered!(Pretending like I am the one who set this problem up, hehehe...
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And thank you for showing me another way to solve this problem too!
How is that? It's not given it's an integer is it?we only need to consider the cases $\sqrt{x}=1$ and $\sqrt{x}=2$ for $x$.
I think math is our true friend, as figures never lie, though humans do!I'm glad to see you enjoy math so much.![]()
Ops! I'm terribly sorry for making a false assumption in my solution, I blindly assumed $\sqrt {x}$ and $\sqrt {x}$ are meant to take integers values, which wasn't correct...How is that? It's not given it's an integer is it?
I'd say you're back to the square root.I'm back to square one...![]()
I would start fromHey anemone!
Nice little puzzle.
Here's my solution.
Rewrite as
$$\frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+\sqrt{y}=42$$
Note that $x,y\ge 0$ for the square root to be defined.
Define:
$$u=\frac{4}{\sqrt{x}}+\sqrt{x},\quad v=\frac{9}{\sqrt{y}}+\sqrt{y}$$
So:
$$9u+v=42 \qquad\qquad (1)$$
Note that $u,v\ge 0$, since the square roots are non-negative.
Rewrite:
\begin{array}{lcl}
(\sqrt{x})^2 - u\sqrt{x} +4=0, &\quad& (\sqrt{y})^2 - v\sqrt{y} + 9=0 \\
\sqrt{x} = \frac 12 \left(u \pm \sqrt{u^2 - 4^2}\right), && \sqrt{y} = \frac 12 \left(v \pm \sqrt{v^2 - 6^2}\right)
\end{array}
It follows that $u \ge 4$ and $v \ge 6$.
Then from (1) we get $u =4$ and $v = 6$.
Therefore:
\begin{array}{llcll}
\sqrt{x} &= \frac 12 \cdot 4= 2,&\quad& \sqrt{y}&=\frac 12 \cdot 6 = 3 \\
x&=4,&& y&=9
\end{array}