SOLUTION: I need to find the domain of 2x/(x+5)^2 so I take (x+5)^2 and make it x^2+10x+25 = 0 that is as far as I got then i would put the 25 over and divide by 10 geting x^2=-2.5

Algebra ->  Functions -> SOLUTION: I need to find the domain of 2x/(x+5)^2 so I take (x+5)^2 and make it x^2+10x+25 = 0 that is as far as I got then i would put the 25 over and divide by 10 geting x^2=-2.5      Log On


   

Question 587194: I need to find the domain of 2x/(x+5)^2 so I take (x+5)^2 and make it x^2+10x+25 = 0 that is as far as I got then i would put the 25 over and divide by 10 geting x^2=-2.5
Answer by Schaman_Dempster(26) About Me  (Show Source):
You can put this solution on YOUR website!
You're almost there.
I can see you are setting the denominator to 0 and you have:
++x%5E2+%2B+10x+%2B+25+=+0+
Using Quadratic formula: x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ for the equation +ax%5E2+%2B+bx+%2B+c+
We have a = 1 , b=10, c =25
x+=+%28-+10+%2B-+sqrt%2810%5E2+-+4%2A1%2A25%29%29%2F%282%2A1%29+
or +x+=+%28-+10+%2B-+sqrt%28100+-+100%29%29%2F2+
or +x+=+-10%2F2+
or x =-5
This shows that denominator becomes 0 for x= 5.
Therefore, we can say that x =-5 can not be in the domain of the given fucntion.
Hence, except for x =-5, all real numbers are the domain.
It is represented as:
x ϵ R - {-5}
or x ϵ (-∞,∞) - {5}

Hope this helps~~