Question

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

Answer 1

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

Answer 2

Solution-

The equation given here,

$\frac{1}{x-1} =\frac{x-2}{2x^2-2}$

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

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

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

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

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

$\Rightarrow x=1 \ and -4$

But, at x=1, $\frac{1}{x-1} \ and \ \frac{x-2}{2x^2-2}$ becomes infinity so it can't be a solution to the equation.

∴ x=1 is the extraneous solution of the equation