(-2-2)=0 (-2+2)= -4
(2-2)=0 (2+2)= 4 so they are not the same because the first one is a negative 4 and the second is a positive 4
I am trying to find the domain for x(x+2)(x+2)/(x-2)(x+2). I canceled the (x+2)'s. The only problem is that my book says the domain cannot equal -2. But I say it cannot equal 2 either (along with -2). (-2-2)(-2+2) is equal to (2-2)(2+2), right? Or what did I do wrong? Thanks!
2 answers
x(x+2)(x+2)/(x-2)(x+2) line 1
= x(x+2)/(x-2) line 2
you are dealing with explicit vs implicit restrictions.
Some authors of textbooks will list as restrictions all values which make any of the denominators zero.
Other authors, and yours is one of them, will restrict only those factors which were divided out
let me explain:
let x=2
line 1 = 32/0 which is undefined
line 2 = 8/0 which is undefined
so technically line 1 is still equal to line 2 when x=2 and no need to restrict it as (x cannot be equal to 2)
let x=-2
line 1 = 0/0 which is indeterminate (wait till you take Calculus for that one)
line 2 = 0/-4 = 0 a real number
so clearly line 1 is no longer equal to line 2, making x = -2 a "necessary" restriction.
It is a fine distinction. I used to accept both types of answers.
= x(x+2)/(x-2) line 2
you are dealing with explicit vs implicit restrictions.
Some authors of textbooks will list as restrictions all values which make any of the denominators zero.
Other authors, and yours is one of them, will restrict only those factors which were divided out
let me explain:
let x=2
line 1 = 32/0 which is undefined
line 2 = 8/0 which is undefined
so technically line 1 is still equal to line 2 when x=2 and no need to restrict it as (x cannot be equal to 2)
let x=-2
line 1 = 0/0 which is indeterminate (wait till you take Calculus for that one)
line 2 = 0/-4 = 0 a real number
so clearly line 1 is no longer equal to line 2, making x = -2 a "necessary" restriction.
It is a fine distinction. I used to accept both types of answers.