Find an ordered triple (x,y,z) of real numbers satisfying x ≤ y ≤ z and the system of equations
√x + √y + √z = 10
x + y + z = 38
√(xy) + √(xz) + √(yz) = 30
or, if there is no such triple, enter the word "none" as your answer.
Hi Destyox!
I don't this it's possible, but I'm not sure.
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc +2ac
And if sqrt(x) = a, sqrt(y) = b, sqrt(z) = c, it doesn't check out.
10^2 != 38 + 60
However, I'm not sure if this is a valid method.
I hope this helped. :)))
=^._.^=
If there's a typo, and that 30 should be 31, then 4, 9, 25 would work.
+75 
