It looks pretty simplistic but for me it's a bit complicated >.<.

So this is line one: $a^2 = b^2$

Next we square root both of them. $\sqrt {a^2} = \sqrt {b^2}$

Next move is the part that I'm not sure of. Is it this? $|a| = |b|$ If yes then this?! $\pm a = \pm b$

Please help me.

  • 1
    $\begingroup$ That can be simplified to $a = b$ or $a = -b$. It's correct. The important thing that you successfully remembered was $ \sqrt {a^2} = |a|$ (same wrt b). Another option is proceeding as Dr.Sonnhard has, below. $\endgroup$ – amWhy Sep 24 '17 at 17:25
  • $\begingroup$ Looks fine. However, it is simpler to consider only one pair of signs:$$a=\pm b$$which should suffice. $\endgroup$ – Simply Beautiful Art Sep 24 '17 at 17:29

note that $$a^2 = b^2 \iff a^2-b^2=0\iff (a+b)(a-b) = 0$$

  • $\begingroup$ I was going to write this as a second option but I forgot :P $\endgroup$ – alienCY Sep 24 '17 at 16:51
  • $\begingroup$ Again, if ever I make an edit to a post of yours, I'm only doing so to help your answer succeed.) You are free of course, to roll back the edit, or edit the edit. $\endgroup$ – amWhy Sep 24 '17 at 17:25

Yes, $\sqrt{x^2}=|x|$ for any real number $x$.

Thus, the equation is equivalent to $|a|=|b|$ and there are two solutions: $a=b$ or $a=-b$.

  • $\begingroup$ $a^2 - b^2 = 0$ and $2ab = 1$ Depending on which I solve first I get more solutions if I choose to solve $a^2 - b^2 = 0$ and then substitute the solution into the other equation. $\endgroup$ – alienCY Sep 24 '17 at 16:53
  • $\begingroup$ @alienCY Why do you think $2ab = 1$? Check your algebra. $\endgroup$ – Ethan Bolker Sep 24 '17 at 17:26
  • $\begingroup$ Trying to solve $\sqrt i$ I think the answer lies on the fact that $a \in R$. I watched this video: youtube.com/watch?v=Z49hXoN4KWg&t=302s $\endgroup$ – alienCY Sep 24 '17 at 17:40

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