SOLUTION: Hi, I have a question about how to solve this equation: 2+x>15/x. This is how I would solve it: 2x+x^2>15 x^2+2x-15>0 (x+5)(x-3)>0 with a result of x>-5 or x<3 So the ans

Question 289050: Hi,
I have a question about how to solve this equation: 2+x>15/x.
This is how I would solve it:
2x+x^2>15
x^2+2x-15>0
(x+5)(x-3)>0
with a result of x>-5 or x<3
So the answer would be (-5,0),(3, infinity)....
Yet, I think I missed something in my steps for solving the equation...because once, when I redid the equation, I found the answer to be (-infinity,-5),(0,3).
Could you please help me find the correct solution?
Thanks
Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
I have a question about how to solve this equation: 2+x>15/x.
This is how I would solve it:
2x+x^2>15
x^2+2x-15>0
(x+5)(x-3)>0
with a result of x>-5 or x<3
So the answer would be (-5,0),(3, infinity)....
Yet, I think I missed something in my steps for solving the equation...because once, when I redid the equation, I found the answer to be (-infinity,-5),(0,3).
GREAT START!
(x+5)(x-3) > 0 when 1.)both factors are positive or when 2.)both factors are negative.
So for situation 1.):
x+5 > 0 AND x-3 > 0
That is when x > -5 AND x > 3 which is the open interval (3, infinity)
For situation 2.):
x+5 < 0 AND x-3 < 0
That is x < -5 AND x < 3 which is the open interval (-infinity, -5)
The total solution set is the union of (-infinity, -5) union (3, infinity).
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