+79 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!
+128474 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
