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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

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- Mar 5, 2012

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Hey 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}

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- Feb 14, 2012

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HiHey anemone!

Nice little puzzle.

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...

And thank you for showing me another way to solve this problem too!

My solution:

We're given to solve for $x$ and $y$ from the equation \(\displaystyle \frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}\).

Here we know $\sqrt{x}, \sqrt{y}>0$.

By rewriting the given equation to get

\(\displaystyle \frac{36}{\sqrt{x}}+9\sqrt{x}+\frac{9}{\sqrt{y}}+ \sqrt{y}=42\)

It's not hard to see that $\sqrt{x}\le2$ because when $\sqrt{x}>2$, e.g.

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 $y$. |

And since $\sqrt{x}, \sqrt{y}>0$ and $\sqrt{x}\le2$, we only need to consider the cases $\sqrt{x}=1$ and $\sqrt{x}=2$ for $x$.

When $\sqrt{x}=1$, we have

\(\displaystyle \frac{36}{1}+9(1)+\frac{9}{\sqrt{y}}+ \sqrt{y}=42\)

\(\displaystyle \frac{9}{\sqrt{y}}+ \sqrt{y}=-3\)

Similarly, this equation has no real number solution.

When $\sqrt{x}=2$, we have

\(\displaystyle \frac{36}{2}+9(2)+\frac{9}{\sqrt{y}}+ \sqrt{y}=42\)

\(\displaystyle \frac{9}{\sqrt{y}}+ \sqrt{y}=6\)

\(\displaystyle \sqrt{y}=3\)

\(\displaystyle \therefore \;\;(x,y)=(4, 9)\).

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- Mar 5, 2012

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I'm glad to see you enjoy math so much.HiI 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...)

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 $x$.

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- Feb 14, 2012

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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?

Sorry! and now, I'm back to square one...

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- Mar 5, 2012

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I'd say you're back to the square root.I'm back to square one...

- Mar 31, 2013

- 1,322

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}

36/x^(1/2) + 9x^(1/2) + 9 /y^(1/2) + y^(1/2) = 42

And proceed as

(36/x^(1/2) + 9x^(1/2) – 36) + (9 /y^(1/2) + y^(1/2)- 6) = 0

Hence 9( 2/x^1/4 – x^(1/4))^2 + (3/y^(1/4) – y^(1/4))^2 = 0

=> 2/x^(1/4) – x^(1/4) = 0 or x = 2^2 = 4

And 3/y^(1/4) – y^(1/4) = 0 or y = 9