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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{xyz} &= 30, \end{align*}or, if there is no such triple, enter the word "none" as your answer.

 Nov 27, 2018
 #1
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\(x\le y\le z\)

\( sqrt{x} + sqrt{y} + sqrt{z} = 10\)

\(x + y + z = 38\)

\(sqrt{xyz} = 30\)

 

Let sqrt x  = a, sqrt y = b  and sqrt z = c...so we have

 

a + b + c  = 10      

a^2 + b^2 + c^2 = 38

abc = 30  ⇒ c =  30 / [ ab]    

 

c is an integer > a, b    

By inspection,  the system  will be true  when a = 2 and b = 3

And c = 30 / [2 * 3 ]  =  5

 

So

The solutions  are  (a, b, c)    = ( 2, 3 , 5)

 

So

(x , y , z)   =  (a^2, b^2,c^2) =    (4, 9, 25)

 

 

cool cool cool

 Nov 27, 2018
edited by CPhill  Nov 27, 2018

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