Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrtx + sqrty +sqrtz = 10
x+y+z=42
(sqrtxyz) = 20
+14995 Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrtx + sqrty +sqrtz = 10
x+y+z=42
(sqrtxyz) = 20
Hello Guest!
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=10\)
\(x+y+z=42\)
\(\sqrt{xyz}=20\)
\(xyz=400\)
\(z=\frac{400}{xy}=42-x-y\)
\(z= (10-\sqrt{x}-\sqrt{y})^2\)
\(I.\ \frac{400}{xy}=42-x-y\)
\(II.\ \frac{400}{xy}=(10-\sqrt{x}-\sqrt{y})^2 \)
\( y\in\{-3,3\}\\ x\in \{-2,2\}\)
\(z\in \{5,9,11,15\}\)
\(z=3\)
\(x+y+3=42\\ x+y=39\)
Error! To be continued.
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