SOLUTION: if the lengths of the sides of a triangle ABC is 3,x+3, and 9 what are all possible values of x?

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Question 720944: if the lengths of the sides of a triangle ABC is 3,x+3, and 9 what are all possible values of x?
Found 2 solutions by Edwin McCravy, checkley79:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
3, x+3, and 9

The three inequalities every triangle must have are

1. first side + second side > third side
2. first side + third side > second side
3. second side + third side > first side

So

1.   3 + x+3 > 9
         6+x > 9
           x > 3

2.   2 + 9 > x+3
        11 > x+3
         8 > x

3.   x+3 + 9 > 3
      x + 12 > 3
           x > -9

All those will be true if 3 < x < 8

Edwin

Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
3,x+3, and 9
3^2+(X+3)^2=9^2
9+X^2+6X+9=81
X^2+6X+18-81=0
X^2+6X-63=0
(X-5.485)(X+11.485)=0
X-5.485
X=5.485 ANS.
--------------------
3^2+(X-3)^2=9^2
9+X^2-6X+9=81
X^2-6X+18=81
X^2-6X+18-81=0
X^2-6X-63=0
(X+5.485)(X-11.485)=0
X-11.485=0
X=11.485 ANS.
---------------------
3^2+9^2=(X+3)^2
9+81=X^2+6X+9
X^2+6X+9-9-81=0
X^2+6X-81=0
(X-6.487)(X+12.487)=0
X-6.487=0
X=6.487 ANS.
-----------------------
3^2+9^2=(X-3)^2
9+81=X^2-6X+9
X^2-6X+9-9-81=0
X^2-6X-81)=0
(X+6.487)(X-12.487)=0
X-12.487=0
X=12.487 ANS.