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Feb 14, 2023 at 5:24 comment added ryang \begin{align} \leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\ ={}& |a|^2+2|a| \cdot|b|+|b|^2. \tag{2} \end{align} - Neither is fragment $(2)$ replacing fragment $(1);$ the symbol is not being dropped and subsequently ignored!
Feb 13, 2023 at 18:56 comment added ryang YES. Your observation in paragraph 3 reminds me of an answer that I posted just now postulating why the student misunderstood the working below; the following excerpt was added only subsequent to my initial submission because this perspective genuinely hadn't occured to me prior:
Feb 13, 2023 at 18:38 history answered kaya3 CC BY-SA 4.0