SOLUTION: Need help determining the restricted values. (x^2-4x-21)/(x^2+16) I think this is the answer. x^2+16=0 subtract 16 to both sides x^2=-16

Algebra ->  Algebra  -> Distributive-associative-commutative-properties -> SOLUTION: Need help determining the restricted values. (x^2-4x-21)/(x^2+16) I think this is the answer. x^2+16=0 subtract 16 to both sides x^2=-16      Log On

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Question 171833: Need help determining the restricted values.
(x^2-4x-21)/(x^2+16)
I think this is the answer.
x^2+16=0 subtract 16 to both sides
x^2=-16
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
You're on the right track. I'll start where you left off.


x%5E2=-16 Start with the given equation


x=sqrt%28-16%29 or x=-sqrt%28-16%29 Take the square root of both sides. (note: don't forget the plus/minus)



Now since you CANNOT take the square root of a negative number, this means that "x" is not a real number. So this means that you can plug in ANY number in for "x" and no division of zero will occur.


So the domain is all real numbers (ie there are no restricted values)