Question

What is the extraneous root of the rational equation `\frac { 1} { x + 1} + \frac { x } { x - 3} = \frac { 12} { x ^ { 2} - 2x - 3}`?

Answer 1

`3` is an extraneous root of the rational equation. [Ans]

Explanation:

`1/(x+1) +x/(x-3) = 12/(x^2-2x-3)` or

`1/(x+1) +x/(x-3) = 12/((x+1)(x-3)) ; x !=3 , x!=-1`
`[ x^2-2x-3=(x+1)(x-3) ]`.

Multiplying by `(x+1)(x-3)` in both sides we get

`(x-3) +x(x+1) =12 or x-3 +x^2 +x -12 =0` or

`x^2+2x-15=0 or (x+5)(x-3) =0`

Roots are `x= -5, x=3`, but `3` is excluded from the domain of

the original equation because it would make the denominator zero

and division by zero is not allowed. Hence `3` is an extraneous

root of the rational equation. [Ans]