Proving $x-8$ is equal to $x^{2}- 64$

Question

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.

$x^2-64$ is not equal to $x-8$ , but you can see that $x^2-64=x^2-8^2=(x-8)(x+8)$ — Bernstein, Jun 03 '18 at 13:15

score 5 · Accepted Answer · answered Jun 03 '18 at 13:13

5

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

answered Jun 03 '18 at 13:13

Green.H

1,201

$x−8=(x+8)(x−8) \Rightarrow 1=(x+8)$ – Ramsha Khalid Jun 03 '18 at 13:16
I cancelled $x-8$ from both sides – Ramsha Khalid Jun 03 '18 at 13:16
1

@RamshaKhalid you cannot do it since $x-8=0.$ You cannot divide by zero. – Green.H Jun 03 '18 at 13:17

Proving $x-8$ is equal to $x^{2}- 64$

1 Answers1