Question

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

Answer 1

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`

Answer 2

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.