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

 Sep 6, 2023

Best Answer 

 #1
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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.

 Sep 6, 2023
 #1
avatar
+1
Best Answer

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.

Guest Sep 6, 2023

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