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when solving $x^6=x^4$ the correct points of intersection occur when solving in this manner:

$x^4(x^2-1)=0, \Rightarrow x=0, x=\pm 1$

So, then why does attempting to solve the same problem via this way:

$x^6=x^4$ then dividing $x^4$ from both sides $\Rightarrow x^2=1 \Rightarrow x=\pm 1$ yield the wrong answer? The algebraic move was valid is it not?

veritas
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    It is valid iff $x \neq 0$, so you have to look at the case $x=0$ separatly! – Stefan Sep 25 '13 at 08:36
  • You divide by $0$. – Caran-d'Ache Sep 25 '13 at 08:37
  • The equation is of the 6th power, so it should have 6 roots. You divide by $x^4$ assuming that 0 is not the root and you get only 2 roots. So where are the 4 of them left? :) – Caran-d'Ache Sep 25 '13 at 08:40
  • There are four roots of $x^4=0$... zero repeated four times. The problem with the language here is that via the factor theorem roots are equivalent to factors and when we say a polynomial of degree $n$ (over the complex numbers to be safe) has $n$ roots we really mean $n$ (linear) factors. – JP McCarthy Sep 25 '13 at 09:28

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Quoting Stefan:

It is valid iff $x\ne 0$, so you have to look at the case $x=0$ separately!

There doesn't seem to be anything more to say.

dfeuer
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