Question

How do you simplify and find the restrictions for `(x^2+3x-18)/(x^2-36)`?

Answer 1

The restricted values of `x` are: `x_1=6` and `x_2=-6`.

Explanation:

There is no simplification possible.

To find the restrictions you have to see for what values of `x` there's no solution. That would be when the denominatior is `0`.

So you get,
`x^2-36=0`

you isolate `x`,
`x^2=36`

and you do the square root in both sides,
`x=+-sqrt(36)=+-6`

so these are the restricted values on x:
`x_1=6`
`x_2=-6`

Answer 2

`f(x)=` `(x^2+3x-18)/(x^2-36)=((x-3)(x+6))/((x-6)(x+6))=(x-3)/(x-6)`

( `f(x)=0` `<=>` `x=3`

• `x≠6 and x≠-6`
  `Df=(-oo,-6)U(-6,6)U(6,+oo)`
  `Df= R-{-6,6}` )

Answer 3

`(x^2+3x-18)/(x^2-36)` simplifies to `(x-3)/(x-6)` with the restriction that `x!=6` and `x!=+6`

Explanation:

`(x^2+3x-18)/(x^2-36)`

`color(white)("XXX")=((x+6)(x-3))/((x+6)(x-6))`

`color(white)("XXX")`Note the division is only defined if `x!=+-6`

`color(white)("XXX")=(x-3)/(x-6)` provided `(x+6)!=0`