Real numbers $x$ and $y$ have an arithmetic mean of $18$ and a geometric mean of $\sqrt{47}$. Find $x^2+y^2$.

Answer 1

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, $(x+y)^2$.