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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.

 Jan 28, 2021
 #1
avatar+819 
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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. :)))

 

=^._.^= 

 Jan 28, 2021
 #2
avatar+186 
0

If there's a typo, and that 30 should be 31, then 4, 9, 25 would work.

 Jan 29, 2021

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