An extraneous solution is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.
Given the equation:
[tex] \frac{1}{x-1} = \frac{x-2}{2x^2-2} \\ \\ 2x^2-2=(x-1)(x-2)=x^2-3x+2 \\ \\ x^2+3x-4=0 \\ \\ (x-1)(x+4)=0 \\ \\ x-1=0 \ or \ x+4=0 \\ \\ x=1 \ or \ x=-4[/tex]
From the equation, it is obtained that x = 1 and x = -4 are the solutions of the equation.
Now, we substitute x = 1 into the equation as follows:
[tex] \frac{1}{1-1} = \frac{1-2}{2(1)^2-2} \\ \\ \frac{1}{0} = \frac{-1}{2-2} \\ \\ \frac{1}{0} = \frac{-1}{0} [/tex]
As can be see, for x = 1, the equation is undefined.
Now, we substitute x = -4 into the equation as follows:
[tex]\frac{1}{-4-1} = \frac{-4-2}{2(-4)^2-2} \\ \\ \frac{1}{-5} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} \\ \\ - \frac{1}{5} =- \frac{1}{5} [/tex]
It can be seen that x = -4 is a valid solution of the orignal equation.
Therefore, x = 1 is an extraneous solution to the equation
[tex]\frac{1}{x-1} = \frac{x-2}{2x^2-2}[/tex]
