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

Rearranging it I get: $1-x^2=x-x^2$, and so $x=1$. But the question Im doing says to find 2 solutions. How would I find the 2nd solution?

Thanks.

  • 4
    Your solution looks perfectly correct: this equation has a unique solution. Are you sure you copied it down correctly?2011-12-27
  • 0
    An equation is termed as "quadratic" if the coefficient of the highest power (2 in this case) is not zero. As the coefficient of the leading term is zero in the given example hence calling it as a quadratic is not correct.2011-12-27
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    I was just consulting a friend on this, and he gave me the following answer (perhaps linking in with Raymond Manzoni's answer?). $$\frac{1}{x^2} - 1 = \frac{1}{x}-1$$ $$1-x^2=x-x^2$$ Let $x=\frac{1}{k}$ and substitute into the equation above. $$1-\frac{1}{k^2}=\frac{1}{k}-\frac{1}{k^2}$$ Simplifying $$k^2-1=k-1$$ $$k^2-k=0$$ $$k(k-1)=0$$ $k=0$ or $k=1$ So $x=\frac{1}{1} = 1$ and $x=\frac{1}{0}$ and so $\frac{1}{x}=0$ and thus $x=\infty$ Are you allowed to do that?2011-12-27
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    If $\infty$ is allowed as an answer (with $1/\infty =0$), it is easily seen to be a solution. But, I rather doubt a pre-calculus/algebra course would admit $\infty$ as a solution. It is far more likely that whoever told you to find 2 solutions is in error.2011-12-27
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    Thanks, but is the working/method above in my above post valid?2011-12-27
  • 2
    I think so; but it is overly complicated (actually, you can't write $x={1\over 0}$; this is undefined). Just say $1/\infty=0$. Then ${1\over \infty^2}-1={1\over\infty}-1\iff0-1=0-1\iff -1=-1$2011-12-27
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    okay, thanks for your help :)2011-12-27

6 Answers 6