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Question Number 211679 by MrGaster last updated on 16/Sep/24

$$$$$$\mathrm{Inverse}\mathrm{root}\mathrm{formula}:$$$$\mathit{x}=\frac{2\mathit{c}}{-\mathit{b}\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}$$$$\left(1\right)\mathrm{The}‘‘\mathrm{antiroot}{\mathrm{formula}}^{″}\mathrm{is}\mathrm{derivedr}$$$$\mathrm{fom}\mathrm{the}\mathrm{abovementionedt}$$$$\mathrm{antiroo}\mathrm{formula}.$$

Commented by mr W last updated on 16/Sep/24

$$rootsfrom$$$${ax}^2+bx+c=0are$$$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$$${ax}^2+bx+c=0\iff c{\left(\frac{1}{x}\right)}^2+b\left(\frac{1}{x}\right)+a=0$$$$\Rightarrow \frac{1}{x}=\frac{-b\pm \sqrt{b^2-4ca}}{2c}$$$$\Rightarrow x=\frac{2c}{-b\pm \sqrt{b^2-4ac}}$$

Answered by Ghisom last updated on 16/Sep/24

$$x=\frac{2c}{-b\pm \sqrt{b^2-4ac}}{}_{}^{}=$$$$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

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