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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

 Oct 25, 2020
 #1
avatar+10583 
+1

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.

indecision  !

 Oct 26, 2020
edited by asinus  Oct 26, 2020
 #2
avatar+31320 
+2

As follows:

 

My statement above that "There are only four possibilities ..." assumes that we are dealing with integers only.  

 Oct 26, 2020
edited by Alan  Oct 26, 2020

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