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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

User Nookaraju
by
7.6k points

2 Answers

3 votes
x=1 would be extraneous as in both sides of original equation, it would be an asymptote 
User YasirAzgar
by
8.3k points
5 votes

Solution-

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

User Nikolay Moskvin
by
7.9k points

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