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Positive real numbers $x,y$ satisfy the equations $x^2 + y^2 = 1$ and $x^4 + y^4= \frac{17}{18}$. Find $xy$.

 Apr 1, 2021

Best Answer 

 #1
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\((x^2+y^2)^2=1^2\)

\(x^4+y^4+2(xy)^2=1\)

\(\frac{17}{18}+2(xy)^2=1\)

\(xy=\sqrt{\frac{1}{36}}=\frac{1}{6}\)

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 Apr 1, 2021
 #1
avatar+144 
+1
Best Answer

\((x^2+y^2)^2=1^2\)

\(x^4+y^4+2(xy)^2=1\)

\(\frac{17}{18}+2(xy)^2=1\)

\(xy=\sqrt{\frac{1}{36}}=\frac{1}{6}\)

Jamesaiden Apr 1, 2021

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