Real numbers $x$ and $y$ have an arithmetic mean of $18$ and a geometric mean of $\sqrt{47}$. Find $x^2+y^2$.
Hi Guest,
If they have an arithmetic mean of 18, this means: \(\dfrac{x+y}{2}=18\)
So, \(x+y=36\)
If they have a geometric mean of \(\sqrt{47}\), this means: \(\sqrt{xy}=\sqrt{47}\)
Then, \(xy=47\)
Now, we want to find \(x^2+y^2\)
We can rewrite it as follows: \((x+y)^2-2xy\) [Check this by expanding it]
So: \(x^2+y^2=(x+y)^2-2xy=(36)^2-2(47)=1296-94=1202\)
I hope this helps.
Hi Guest,
If they have an arithmetic mean of 18, this means: \(\dfrac{x+y}{2}=18\)
So, \(x+y=36\)
If they have a geometric mean of \(\sqrt{47}\), this means: \(\sqrt{xy}=\sqrt{47}\)
Then, \(xy=47\)
Now, we want to find \(x^2+y^2\)
We can rewrite it as follows: \((x+y)^2-2xy\) [Check this by expanding it]
So: \(x^2+y^2=(x+y)^2-2xy=(36)^2-2(47)=1296-94=1202\)
I hope this helps.
