How do you find the restricted values of x or the rational expression #(x^2+x+15)/(x^2-3x)#?

2 Answers
Feb 27, 2016

Answer:

Restrictions on #x# are given by #x!=0# and #x!=3#.

Explanation:

The rational expression #(x^2+x+15)/(x^2−3x)#, cannot be defined if denominator i.e. #x^2-3x=0# i.e. #x(x-3)=0#

i.e. #x# cannot take values #{0,3}#.

Hence, restrictions on #x# are given by #x!=0# and #x!=3#

The values of #x# that the rational expression has a meaning are all the reals that dont nullify the denominator hence

#x^2-3x=0=>x(x-3)=0=>x=0 or x=3#

Hence the domain is #R-{0,3}# where #R# is the set of reals.