+501 Positive real numbers $x,y$ satisfy the equations $x^2 + y^2 = 1$ and $x^4 + y^4= \frac{17}{18}$. Find $xy$.
+150 \((x^2+y^2)^2=1^2\)
\(x^4+y^4+2(xy)^2=1\)
\(\frac{17}{18}+2(xy)^2=1\)
\(xy=\sqrt{\frac{1}{36}}=\frac{1}{6}\)
.+150 \((x^2+y^2)^2=1^2\)
\(x^4+y^4+2(xy)^2=1\)
\(\frac{17}{18}+2(xy)^2=1\)
\(xy=\sqrt{\frac{1}{36}}=\frac{1}{6}\)
