# Find Real Solution(s) Challenge

#### anemone

##### MHB POTW Director
Staff member
Find the real solution(s) to the equation $$\displaystyle \frac{36}{\sqrt{x}}+\frac{9}{\sqrt{y}}=42-9\sqrt{x}-\sqrt{y}$$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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!

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}

#### anemone

##### MHB POTW Director
Staff member
Hey anemone!

Nice little puzzle.
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... )

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

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

$$\displaystyle \frac{9}{\sqrt{y}}+ \sqrt{y}=6$$

$$\displaystyle \sqrt{y}=3$$

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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... )

And thank you for showing me another way to solve this problem too!
I'm glad to see you enjoy math so much. we only need to consider the cases $\sqrt{x}=1$ and $\sqrt{x}=2$ for $x$.
How is that? It's not given it's an integer is it?

#### anemone

##### MHB POTW Director
Staff member
I'm glad to see you enjoy math so much. I think math is our true friend, as figures never lie, though humans do! How is that? It's not given it's an integer is it?
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...

Sorry! and now, I'm back to square one... #### Klaas van Aarsen

##### MHB Seeker
Staff member
I'm back to square one... I'd say you're back to the square root. ##### Well-known member
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}
I would start from

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