Answer:
Therefore x = 1 is an extraneous solution.
Step-by-step explanation:
The solution to the equation StartFraction 1 Over x minus 1 EndFraction = StartFraction x minus 2 Over 2 x squared minus 2 EndFraction
[tex]\Rightarrow \frac{1}{x - 1} = \frac{x- 2}{2x^2-2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 2x^2 - 2 = (x-1)(x-2)[/tex]
therefore [tex]2x^2 - 2 = x^2 - 3x + 2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x^2 + 3x - 4 = 0[/tex]
therefore [tex]x^2 + 3x - 4 = 0 \hspace{0.3cm} \Rightarrow = \hspace{0.3cm} (x-1)(x+4) = 0 \\[/tex]
therefore x = 1 and x = -4 but we can see that if we place x = 1 in the original solution then [tex]\frac{1}{x-1} = \frac{1}{1-1} = \frac{1}{0}[/tex] which is indeterminate.
Therefore x = 1 is an extraneous solution.
Answer:
C. x = 1
Step-by-step explanation:
Which solution to the equation StartFraction 1 Over x minus 1 EndFraction = StartFraction x minus 2 Over 2 x squared minus 2 EndFraction is extraneous?
x = 1 and x = –4
neither x = 1 or x = –4
x = 1
x = –4
