Question

Which solution to the equation 1/x-1 = x-2/2x^2-2 is extraneous

x = 1 and x = –4
neither x = 1 or x = –4
x = 1
x = –4

Answer 1

x=1 would be extraneous as in both sides of original equation, it would be an asymptote 

Answer 2

The equation given here,

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

`\Rightarrow 2x^2-2=(x-1)(x-2)`

`\Rightarrow 2x^2-2=x^2-3x+2`

`\Rightarrow x^2+3x-4=0`

`\Rightarrow x^2 +4x-x-4=0`

`\Rightarrow (x-1)(x+4)=0`

`\Rightarrow x=1 \ and -4`

But, at x=1,
`(1)/(x-1) \ and \ (x-2)/(2x^2-2)` becomes infinity so it can't be a solution to the equation.

∴ x=1 is the extraneous solution of the equation