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7

Here's why you should take great care when considering $\pm$ as an operator. It's not unusual to see a sentence of the form We deduce that $A=\pm B$ and hence that $C=D\pm E$. This isn't simply saying that both ($A$ is either $B$ or $-B$) and (either $C=D-E$ or $C=D+E$). When two $\pm$'s appear in the same sentence it is implied that they are both to be ...


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When solving a quadratic equation, $$ax^2+ bx+ c = 0$$ we use shorthand for the two solutions, to include both $$x_1 = \frac {-b +\sqrt{b^2-4ac}}{2a}$$ and $$x_2 = \frac{-b - \sqrt{b^2 -4ac}}{2a}$$ Hence, the shorthand, $$x_i = \frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$ I.e., the solutions to $ax^2+bx+c = 0$ are given by $$x\in \left\{ \frac {-b +\sqrt{b^2-4ac}...


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