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Find an ordered triple \((x,y,z) \) of real numbers satisfying \(x\le y\le z\) and the system of equations

\(\begin{align*} \sqrt{x} + \sqrt{y} + \sqrt{z} &= 10, \\ x + y + z &= 38, \\ \sqrt{xy} + \sqrt{xz} + \sqrt{yz} &= 30, \end{align*}\)
or, if there is no such triple, enter the word "none" as your answer.

 

Thank you!

 Mar 13, 2021
 #1
avatar+118069 
+2

I've solved something like  this  before with a tedious  (probably  unduly so) process

 

No solutions exist UNLESS we make ONE slight change to equation  3  (letting it equal 31  instead of 30)

 

We  can  then solve this  by  inspection

 

Let   z = 25     y =  9   and  x = 4

 

So

 

sqrt ( 4)  + sqrt ( 9)  + sqrt (25)   =    2  + 3  + 5   =   10

 

And

 

4  +  9  +  25  =   38

 

And

 

sqrt (4*9)  + sqrt (4 *25)  +  sqrt (9 * 25)   =

 

sqrt (36)  + sqrt (100) +  sqrt (225)  =

 

6    +   10    +    15  =     31

 

 

cool cool cool

 Mar 13, 2021

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