SOLUTION: Find an ordered triple (x,y,z) of real numbers satisfying the system of equations sqrt{x} + sqrt{y} + sqrt{z} = 10, x + y + z = 38, sqrt{xyz} = 30, or, if there is no such

Let me introduce new variables u = , v = ,  and w = .

Then I can rewrite the given equations in this form:

u + v + w = 10,   (1)

 = 38,    (2)

u*v*w = 30.       (3)


From (1) and (2)

 = 100 =  =  = 38 + 2(uv + uw + vw),

hence,  2(uv + uw + vw) = 100 - 38 = 62,   and  uv + uw + vw = 31.


Now I can rewrite the system (1),(2),(3) in the form

u + v + w = 10,       (1')

uv + uw + vw = 31,    (2')

u*v*w = 30.           (3')


It means ( !! The Vieta's formulas !! ) that u, v ans w are the roots of this cubical equation

X^3 - 10X^2 + 31X - 30 = 0.   (4)


Now you can apply the "Rational roots Theorem" which says that the rational (in particular, integer) roots of the equation (4) 
are among the divisors of the constant term 30.

Or even better, you can make a plot of the left side polynomial which (the plot) will say you a lot about the roots:





Plot y = 


From the plot, it is clearly seen that the roots u, v and w are 2, 3 and 5 (from smaller to greater).


You can check it manually.


Hence, the solution to the original system is (x,y,z) = (4,9,25) and all possible permutations.