+15 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.
+128448 \(x\le y\le z\)
\( sqrt{x} + sqrt{y} + sqrt{z} = 10\)
\(x + y + z = 38\)
\(sqrt{xyz} = 30\)
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)
