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How this process has been calculated in a manner added.

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marked as duplicate by Rahul, Hakim, Will Jagy, Mark Bennet, Michael Albanese Sep 26 '14 at 19:09

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  • $\begingroup$ The mistake is already in the second step: $$4-\frac92\neq\sqrt{\left(4-\frac92\right)^2}$$ since $\;\sqrt{x^2}=|x|\;$ , and in this case $\;4-\frac 92<0\;$ . $\endgroup$ – Timbuc Sep 26 '14 at 18:29
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Note that $$4-\frac 92\not=\sqrt{\left(4-\frac 92\right)^2}$$

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$$\sqrt{\left(4-\frac92\right)^2}=-\left(4-\frac92\right)$$

More generally for real $a,$ $$\sqrt{a^2}=|a|=\begin{cases} a &\mbox{if } a\ge0 \\ -a & \mbox{if } a<0\end{cases}$$

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