SOLUTION: Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations sqrt{x} + sqrt{y} + sqrt{z} = 10, x + y + z = 38, sqrt{xy} + sqrt{xz} + sqrt{y

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Question 1175809: Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrt{x} + sqrt{y} + sqrt{z} = 10,
x + y + z = 38,
sqrt{xy} + sqrt{xz} + sqrt{yz}= 30,
or, if there is no such triple, enter the word "none" as your answer.

Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52756)   (Show Source): You can put this solution on YOUR website!
,
Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrt{x} + sqrt{y} + sqrt{z} = 10,
x + y + z = 38,
sqrt{xy} + sqrt{xz} + sqrt{yz}= 30,
or, if there is no such triple, enter the word "none" as your answer.
~~~~~~~~~~~~

To simplify my writhing, I introduce new variables


    a = ;  b = ;  c = .


Then from the given conditions, I have


    a + b + c = 10           (1)

    a^2 + b^2 + c^2 = 38     (2)

    ab + ac + bc = 30.       (3)


Square equation (1)  and use this identity  (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.

You will get then


    10^2 = 38 + 2*30,   or

    100  = 38 + 60 = 98.     (4)


Since equality (4) is IMPOSSIBLE, it means that the given system of equations HAS NO solution.

Proved and solved.

Explained and completed.



Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrt{x} + sqrt{y} + sqrt{z} = 10,
x + y + z = 38,
sqrt{xy} + sqrt{xz} + sqrt{yz}= 30
======================
(4,9,25) IF sqrt{xy} + sqrt{xz} + sqrt{yz}= 31

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