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Inverse-root-formula-x-2c-b-b-2-4ac-1-The-antiroot-formula-is-derivedr-fom-the-abovementionedt-antiroo-formula-




Question Number 211679 by MrGaster last updated on 16/Sep/24
             Inverse root formula:            x=((2c)/(−b±(√(b^2 −4ac))))  (1)The“ antiroot formula” is derivedr  fom the abovementionedt  antiroo formula.
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{Inverse}\:\mathrm{root}\:\mathrm{formula}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{c}}}{−\boldsymbol{{b}}\pm\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{ac}}}} \\ $$$$\left(\mathrm{1}\right)\mathrm{The}“\:\mathrm{antiroot}\:\mathrm{formula}''\:\mathrm{is}\:\mathrm{derivedr} \\ $$$$\mathrm{fom}\:\mathrm{the}\:\mathrm{abovementionedt} \\ $$$$\mathrm{antiroo}\:\mathrm{formula}. \\ $$
Commented by mr W last updated on 16/Sep/24
roots from  ax^2 +bx+c=0 are  x=((−b±(√(b^2 −4ac)))/(2a))  ax^2 +bx+c=0 ⇔c((1/x))^2 +b((1/x))+a=0  ⇒(1/x)=((−b±(√(b^2 −4ca)))/(2c))  ⇒x=((2c)/(−b±(√(b^2 −4ac))))
$${roots}\:{from} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{are} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\Leftrightarrow{c}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +{b}\left(\frac{\mathrm{1}}{{x}}\right)+{a}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ca}}}{\mathrm{2}{c}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$
Answered by Ghisom last updated on 16/Sep/24
x=((2c)/(−b±(√(b^2 −4ac)))) _^ =  =((−b±(√(b^2 −4ac)))/(2a))
$${x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\:_{} ^{} = \\ $$$$=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$

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