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 Three numbers (Posted on 2003-11-17)
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have ?

 No Solution Yet Submitted by Ravi Raja Rating: 3.7500 (4 votes)

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 solution | Comment 4 of 21 |
(Corrected per SilverKnight's comment (#8)):

Substituting z=5-x-y into xy+yz+xz=3 gives
xy+y(5-x-y)+x(5-x-y)=3

which can be simplified to
-y²+(5-x)y+5x-x²-3=0

Considered as an equation to be solved for y, but without actually solving, we get a discriminant of 13+10x-3x² This must be positive in order for there to be a real solution for y.

The bounds of where this is positive are those two points where it is zero. So, setting that discriminant to zero, we get

x=(10±√256)/6 = (10±16)/6
or x = 13/3 or -1

Between these two values of x, the discriminant is positive, so y has a real value (as well as z).

So in answer to the question, the largest value that x can have is 13/3.
Edited on November 18, 2003, 9:12 am
 Posted by Charlie on 2003-11-17 15:31:51

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