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# Find an ordered triple (x,y,z) of real numbers

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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
+10583
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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

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.

!

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