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This question already has an answer here:

Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it?

                      -20 = -20
                    16-36 = 25-45
             16-36+(81/4) = 25-45+(81/4)
(4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2)
              (4-(9/2))^2 = (5-(9/2))^2
                  4-(9/2) = 5-(9/2)
                        4 = 5
                      2+2 = 5
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marked as duplicate by user21820, Jonas Meyer, user91500, Claude Leibovici, Chris Godsil Apr 20 '17 at 12:25

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    $\begingroup$ $x^2=y^2$ does not necessarily imply that $x=y$. Not to meantion $(5^2)-(2*5*(9/2))+((9/2)^2)\neq (5+(9/2))^2$ anyways. $\endgroup$ – user12345 Apr 19 '17 at 16:35
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    $\begingroup$ You've gotten $(4-\frac92)^2=(5-\frac 92)^2$ which isn't strange $\endgroup$ – kingW3 Apr 19 '17 at 16:35
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    $\begingroup$ Taking roots on both sides of $(4-(9/2))^2=(5-(9/2))^2$ you get $|4-(9/2)|=|5-(9/2)|$. The rest should be obvious. $\endgroup$ – mlc Apr 19 '17 at 16:38
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    $\begingroup$ Also please consider learning to typeset using MathJax. $\endgroup$ – mathreadler Apr 19 '17 at 16:43
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    $\begingroup$ It's as if you wrote $(-2)^2=(2)^2\implies{}$$-2=2$. $\endgroup$ – Akiva Weinberger Apr 19 '17 at 17:14
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Hint -

$a^2-2ab+b^2=(a-b)^2\ne(a+b)^2$

This is your mistake.

Also $x^2=y^2$ doesn't imply that $x=y$ is the only possibility.

It should be $|x|=|y|$.

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  • $\begingroup$ I corrected it, still, that's not what I want $\endgroup$ – Deepesh Choudhary Apr 19 '17 at 16:40
  • $\begingroup$ What if we take square roots on both sides? $\endgroup$ – Deepesh Choudhary Apr 19 '17 at 16:46
  • $\begingroup$ There is difference in things. Suppose $x=\sqrt9$ then $x=3$. If $x^2=9$ then $x=\pm3$. $\endgroup$ – Kanwaljit Singh Apr 19 '17 at 16:48
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    $\begingroup$ Deepesh $\sqrt{x^2} = |x| \neq x$, as Kanwaljit notes above, $\endgroup$ – Namaste Apr 19 '17 at 16:48
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    $\begingroup$ Your answer is inaccurate. By saying "doesn't mean that $x=y$ only", you are actually implying "not only $x=y$ but something else as well". Instead you should say "doesn't necessarily imply $x=y$" or "doesn't imply that $x=y$ is the only possibility". $\endgroup$ – user21820 Apr 19 '17 at 17:39
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The problem is when you go from a quadratic to a linear.

$$(4-\frac{9}{2})^2 = (5-\frac{9}{2})^2$$ $$|4-\frac{9}{2}| = |5-\frac{9}{2}|$$ $$|\frac{-1}{2}| = |\frac{1}{2}|$$ $$\frac{1}{2} = \frac{1}{2}$$

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$$4-\frac92=\pm\left(5-\frac92\right)$$ so that

$$4=5\lor4+5=\frac92+\frac92$$ and the world is safe...

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Hint :

$$a^2=b^2 \implies |a|=|b|$$

That is :

$$a=b ~~\text{or}~~ a=-b$$

Therefore $$\Big(4-\frac{9}{2}\Big)^2=\Big(5-\frac{9}{2}\Big)^2$$ $$\implies \Big|\Big(4-\frac{9}{2}\Big)\Big|=\Big| \Big(5-\frac{9}{2}\Big) \Big|$$ $$\implies 0.5=0.5$$ Which is absolutely correct.

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