Question

How do you solve `(x+2)/(x+3)-1=1/(3-2x-x^2)` and check for extraneous solutions?

Answer 1

Please see the explanation.

Explanation:

Given: `(x+2)/(x+3)-1=1/(3-2x-x^2)`

Multiply the right side by 1 in the form of `(-1)/(-1)`:

`(x+2)/(x+3)-1=-1/(x^2+2x-3)`

Factor the right side's denominator:

`(x+2)/(x+3)-1=-1/((x-1)(x+3))`

Restrict the domain to prevent division by 0:

`(x+2)/(x+3)-1=-1/((x-1)(x+3)); x!=1, x!=-3`

NOTE: We restrict the values of x to make us notice, if we obtain an extraneous root.

Multiply both sides by `x+3`

`(x+2)-x - 3=-1/(x-1); x!=1, x!=-3`

Combine like terms:

`-1 = -1/(x-1); x!=1, x!=-3`

Multiply both sides by `-(x-1)`:

`x-1 = 1; x!=1, x!=-3`

Add 1 to both sides and drop the restrictions, because the solution will not violate them.

`x = 2`

Check:

`(2+2)/(2+3)-1=1/(3-2(2)-2^2)`

`4/5-1 = 1/-5`

`-1/5=-1/5`

This checks.