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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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4  
Your solution looks perfectly correct: this equation has a unique solution. Are you sure you copied it down correctly? –  Srivatsan Dec 27 '11 at 10:31
    
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. –  user21830 Dec 27 '11 at 10:46
    
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? –  Thomas Dec 27 '11 at 11:34
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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. –  David Mitra Dec 27 '11 at 11:53
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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 ${1\over \infty^2}-1={1\over\infty}-1\iff0-1=0-1\iff -1=-1$ –  David Mitra Dec 27 '11 at 12:11

6 Answers 6

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

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

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.

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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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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) –  Thomas Dec 27 '11 at 11:35

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

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

$$x^2-x=0$$

$$x(x-1)=0$$

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

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1  
$x=0$ cannot be a solution. –  matt Dec 27 '11 at 10:36
    
You multiplied both sides by $x^3$ which introduced the non-solution $x=0$. –  Zev Chonoles Dec 27 '11 at 10:36
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$x=0$ cannot be a solution since one cannot divide by zero. –  E.O. Dec 27 '11 at 10:36
    
@matt: From $x^2=x$ we can get the result $x=0$, but it's not in the domain of the function. –  Gigili Dec 27 '11 at 10:37
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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 $a,c$). –  Marc van Leeuwen Dec 27 '11 at 13:08

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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1  
Your solution is correct, but it is slightly more complicated than necessary: you unnecessarily multiplied both sides by $x^3$. If you multiply both sides of $\frac{1}{x^2}=\frac{1}{x}$ by $x^2$, you get $1=x$, and we don't have to check that $x=0$ isn't a solution because if it were a solution before multiplying both sides by $x^2$, it certainly would be afterwards. –  Zev Chonoles Dec 27 '11 at 10:39
    
@ZevChonoles I did, but I just wanted to point out the possibility that those who made the question might have done the same, making them believe there were 2 solutions. –  E.O. Dec 27 '11 at 10:41
    
Ah, that's a good point. –  Zev Chonoles Dec 27 '11 at 10:42
    
@ZevChonoles: What's the difference between my solution and this one which is correct? I did exactly the same without the explanation of last line . –  Gigili Dec 27 '11 at 10:43
    
@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. –  Zev Chonoles Dec 27 '11 at 10:46

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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