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Okay so I have this trick question that I can't seem to figure out. Suppose $x = 8$ and $x^2 = 64$, then $x -8 = 0$ and
$ x^2- 64 = 0 $. Now comparing both since they are equal to zero, $x - 8 = x^2 - 64 \Rightarrow x - 8 = (x+8) (x-8) \Rightarrow 1 = (x + 8)$, Substituting value of $x$ gives, $1 = 16$ which is not true. I'd appreciate if someone can help me figure out the mistake here.

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    $\begingroup$ $x^2-64$ is not equal to $x-8$ , but you can see that $x^2-64=x^2-8^2=(x-8)(x+8)$ $\endgroup$
    – Bernstein
    Jun 3, 2018 at 13:15

1 Answer 1

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You divided both sides of the equation by $0 = x-8,$ which you cannot do.

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  • $\begingroup$ $x−8=(x+8)(x−8) \Rightarrow 1=(x+8)$ $\endgroup$ Jun 3, 2018 at 13:16
  • $\begingroup$ I cancelled $x-8$ from both sides $\endgroup$ Jun 3, 2018 at 13:16
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    $\begingroup$ @RamshaKhalid you cannot do it since $x-8=0.$ You cannot divide by zero. $\endgroup$
    – Green.H
    Jun 3, 2018 at 13:17

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