SOLUTION: A triangle's sides have lengths 1/(x+2), 1/(x+1) and 1/x what are the possible values of x? a) 0 < x < square root 2 b) x > 0 c) x > 1 d) x< 1 e) x > square root 2

Algebra ->  Algebra  -> Inequalities -> SOLUTION: A triangle's sides have lengths 1/(x+2), 1/(x+1) and 1/x what are the possible values of x? a) 0 < x < square root 2 b) x > 0 c) x > 1 d) x< 1 e) x > square root 2      Log On

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Question 232388: A triangle's sides have lengths 1/(x+2), 1/(x+1) and 1/x what are the possible values of x?
a) 0 < x < square root 2 b) x > 0 c) x > 1 d) x< 1 e) x > square root 2
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
x>0 will do, since the sides of the triangle cannot be negative. In addition, the values of x must NEVER cause a denominator to be zero, which means that ALSO, x cannot be 0, -1, or -2. However, if you select x>0, you have eliminated these problems as well.

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida