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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.

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    okay, thanks for your help :)2011-12-27

6 Answers 6

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There is only one answer which is $x=1$ as you said. It might be possible however that the book erroneously proceeded with the following steps thus leading them to believe that there were 2 answers.

${1\over x^2}-1={1\over x}-1$ ${1\over x^2}={1\over x}$ $x^2=x$ $x^2-x=0$ $x(x-1)=0$ $x=1, x=0$ However by plugging in $0$ into the original equation we get ${1\over 0}-1={1\over 0}-1$. We can therefore discard 0 which leaves us only with 1.

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    @Gigili: The last line is what makes Emile's answer correct. Your answer *incorrectly* states that $x=0$ is a solution without later explaining that is actually *not* a solution.2011-12-27
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The only other solution I can think of is infinity (working in the '(projective) extended real number line')

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    Thanks, could you take a look at my comment above? Am I allowed to work it like that? (see my comment in my original question)2011-12-27
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Another way to see it is

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

Where the case $x=0$ has be excluded manually. Your solution is correct.

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I think it should be emphasised what the salient point is here:

Given the equation $ \Phi =\Psi $ you may multiply both sides by the same non-zero number $a$ to obtain the equivalent equation $ a\Phi =a\Psi. $ Multiplying both sides of an equation by 0 may give an equation that's not equivalent to the original equation.

With your equation, eventually you'll get to the point where you have $ \tag{1}{1\over x^2}= {1\over x}. $ At this point, if you want to "cancel the $x$'s", you could multiply both sides by $x^2$ as long as $x^2\ne0$. You need to consider what happens when $x=0$ separately.

$x=0$ is not a solution of (1) in this case, so the solutions of (1) are the non-zero solutions of $ 1=x. $ If you multiplied both sides of (1) by $x^3$, the solutions would be the non-zero solutions of $ x=x^2. $

Your text made an error, most probably, at this stage...

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

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

$x^2-x=0$

$x(x-1)=0$

The solutions are $x=1$ or $x=0$.

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    @Gigili: The method of transforming the equation $\frac ab=\frac cd$ to $ad=bc$ consists of multyplying both sides by the product $bd$ of the denominators, even if you may not be aware of this. If the denominators are non constant, then this multiplication will in general introduce spurious solutions for which $b=d=0$, which is of course forbidden in the original problem; in the current case it introduces $x=0$. The method *may* also introduce spurious solutions where only one of $b,d$ is zero (depending on the expressions $a,c$).2011-12-27
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$\frac{1}{x^2} - 1 = \frac{1}{x} -1$

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

multiply by x suppose that $x \ne0$

$x^2-x=0$

$x(x-1)=0$

The unique solution is $x=1$ solution $x=0$ is excluded.