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Whenever we multiply an equation by an expression involving a variable, the zero of the expression becomes a solution of the equation. This is natural, because if we have a function $f(x), f(b) \neq a$. We want to solve

$$f(x) = a$$

If we multiply both sides by some expression $g(x)$ such that $g(b) = 0$

$$f(x)\cdot g(x) = a\cdot g(x)$$

Clearly, $x=b$ is a solution of the modified equation but not of the original one.

However, this does not seem to work for equations like

$$6x^{0.75} = 7x^{0.25} - 2x^{-0.25}$$

Multiplying both sides by $x^{0.25}$, we would expect $x=0$ to be a solution. But the equation we get

$$6x = 7\sqrt{x} - 2$$

clearly does not satisfy $x=0$. How come?

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  • $\begingroup$ well, I guess you solve the equation with the condition g(x)≠0, so in that case: x=0 is a solution, then, if x≠0 you multiply by g(x) and find other solutions $\endgroup$
    – Mosk
    Oct 2, 2014 at 15:21
  • $\begingroup$ $x^{-0.25} = \dfrac{1}{x^{0.25}}$, so I don't think $x= 0$ satisfies the given equation. $\endgroup$
    – taninamdar
    Oct 2, 2014 at 15:34

4 Answers 4

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Let us show one more step in your example: $$6x^{0.75} = 7x^{0.25} - 2x^{-0.25}\\ 6x^{0.75}x^{0.25} = 7x^{0.25}x^{0.25} - 2x^{-0.25}x^{0.25}$$

Now you said $x^{-0.25}x^{0.25}=1,$ but that is not true for $x=0$. In fact, your original equation should come with a note $x \neq 0$ because of the $x^{-0.25}$ term.

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In your second example you create an indeterminate form, 0/0, which is why it fails to work.

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You can see the same sort of thing in the simpler situation of solving $x=\frac1x$. Multiplying by $x$ gives $x^2=1$ (at least formally, if we don't concern ourselves too much about $x=0$). And even though we multiplied by $x$, we see that $0$ is not a solution of the resulting equation.

I would say that the reason is that we multiplied something infinity-like at $0$ (namely $\frac1x$) by $x$, which is $0$ at $0$. But our indeterminate $0\times\infty$ worked out to $1$ instead of $0$.

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The original equation isn't defined at $x = 0$.

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