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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} + \

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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+0 Answers

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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)   Nov 27, 2018
edited by CPhill  Nov 27, 2018