Real numbers $x$ and $y$ have an arithmetic mean of $18$ and a geometric mean of $\sqrt{47}$. Find $x^2+y^2$.
Since $x$ and $y$ have an arithmetic mean of $18$, we have that $\frac{x+y}{2} = 18$. From this, we can represent $y$ in terms of $x$ as $y = 36-x$.
Since $\sqrt{47}$ is the geometric mean of $x$ and $y$, we know that $\sqrt{47} = \sqrt{xy} = \sqrt{x(36-x)}$.
Squaring both sides to eliminate the square root sign gives us
$47 = x(36-x)$
This is a quadratic equation, which simplifies to $x^2-36x+47 = 0$.
Factor the quadratic to get $(x-23)(x-2) = 0$
This gives us $x=23$ or $x = 2$
If $x=23$, then $y=36-23 = 13$. If $x=2$, then $y=36-2=34$. This gives us two pairs $(x,y)$ that satisfy the given conditions:
$(23,13)$
$(2,34)$
Finally, the problem asks for the sum of the squares of $x$ and $y$.
$(23^2 + 13^2) = (23+13)(23-13) = (36)(10) = \boxed{360}$
and
$(2^2 + 34^2) = (2+34)(34-2) = (36)(32) = \boxed{1152}$.
Notice that the sums of the squares of $x$ and $y$ for both pairs are actually \textbf{both} equal to the same quantity, $(x+y)^2$.
Let's start by solving for $x$ and $y$ using the information given.
We can use the formulas for arithmetic and geometric means to set up two equations:
Arithmetic mean: $\frac{x+y}{2} = 18$
Geometric mean: $\sqrt{xy} = \sqrt{47}$
To solve for $x$ and $y$, we can solve these equations simultaneously.
First, let's solve the arithmetic mean equation for $x+y$:
$x+y = 2 \cdot 18 = 36$
Now let's solve the geometric mean equation for $xy$:
$xy = (\sqrt{47})^2 = 47$
We now have a system of equations:
$x+y = 36$
$xy = 47$
We can solve this system by substitution or using a quadratic equation. Let's use substitution:
Solve the first equation for $x$:
$x = 36 - y$
Substitute this expression for $x$ into the second equation:
$(36 - y)y = 47$
Expanding and rearranging the equation, we get:
$36y - y^2 = 47$
Rearrange the equation:
$y^2 - 36y + 47 = 0$
This is a quadratic equation, and we can solve it using the quadratic formula:
$y = \frac{-(-36) \pm \sqrt{(-36)^2 - 4(1)(47)}}{2(1)}$
Simplifying,
$y = \frac{36 \pm \sqrt{1296 - 188}}{2}$
$y = \frac{36 \pm \sqrt{1108}}{2}$
$y = \frac{36 \pm 2\sqrt{277}}{2}$
$y = 18 \pm \sqrt{277}$
Therefore, the possible values for $y$ are $18 + \sqrt{277}$ and $18 - \sqrt{277}$.
Using the expression for $x$ we found earlier ($x = 36 - y$), we can find the corresponding values for $x$:
If $y = 18 + \sqrt{277}$, then $x = 36 - (18 + \sqrt{277}) = 18 - \sqrt{277}$
If $y = 18 - \sqrt{277}$, then $x = 36 - (18 - \sqrt{277}) = 18 + \sqrt{277}$
So, the possible pairs of $x$ and $y$ are:
$(x,y) = (18 - \sqrt{277}, 18 + \sqrt{277})$
$(x,y) = (18 + \sqrt{277}, 18 - \sqrt{277})$
Finally, we can find $x^2 + y^2$ for each pair:
For the first pair:
$(18 - \sqrt{277})^2 + (18 + \sqrt{277})^2 = 1296$
For the second pair:
$(18 + \sqrt{277})^2 + (18 - \sqrt{277})^2 = 1296$
So, $x^2 + y^2 = \boxed{1296}$.