What are the excluded values for #(-n)/(n^2-49)#?

2 Answers
Jul 5, 2017

Answer:

See a solution process below:

Explanation:

The excluded values for this problem are for when the denominator is equal to #0#. We cannot divide by #0#.

Therefore, to find the excluded values we need to equate the denominator to #0# and solve for #n#:

#n^2 - 49 = 0#

First, add #color(red)(49)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#n^2 - 49 + color(red)(49) = 0 + color(red)(49)#

#n^2 - 0 = 49#

#n^2 = 49#

Next, take the square root of each side of the equation to solve for #n# while keeping the equation balanced. Remember, taking the square root of a number produces a positive and negative result:

#sqrt(n^2) = +-sqrt(49)#

#n = +-7#

There excluded values are:

#n = -7# and #n = 7#

Jul 5, 2017

Answer:

#n !=2 and x!=-2#

Explanation:

Excluded values in this case are those which will make the denominator equal to #0#. Division by zero is undefined.

#(-n)/(n^2-49)" "larr# factorise the denominator

#(-n)/((n+2)(n-2))#

Neither bracket may be equal to #0#.

#n+2 != 0 rarr n !=-2#

#n-2 != 0 rarr n!= 2#

These are the excluded values.#