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Question Number 93730 by Rio Michael last updated on 14/May/20

what are the reasons for not using x = ((2c)/(−b ±(√(b^2 −4ac)))) as the quadratic formula? i proved it.

$$\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{reasons}\:\mathrm{for}\:\mathrm{not}\:\mathrm{using}\: \\ $$$$\:{x}\:=\:\frac{\mathrm{2}{c}}{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\:\:\mathrm{as}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{formula}?\: \\ $$$$\mathrm{i}\:\mathrm{proved}\:\mathrm{it}. \\ $$

Commented by Rasheed.Sindhi last updated on 14/May/20

((2c)/(−b ±(√(b^2 −4ac)))) =((2c)/(−b ±(√(b^2 −4ac))))×((−b ∓(√(b^2 −4ac)))/(−b ∓(√(b^2 −4ac)))) =((2c(−b ∓(√(b^2 −4ac))))/(b^2 −(b^2 −4ac))) =((2c(−b ∓(√(b^2 −4ac))))/(4ac)) =((−b ∓(√(b^2 −4ac)))/(2a)) =((−b ±(√(b^2 −4ac)))/(2a)) ∴ Your formula is equivalent to the quadratic formula^• But the quadratic formula^• is better simplified form because it doesn′t contain radical in denominator. ^• Here ′the quadratic formula′ means: x=((−b ±(√(b^2 −4ac)))/(2a))

$$\frac{\mathrm{2}{c}}{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{c}}{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}×\frac{−{b}\:\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{−{b}\:\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$$$\:\:=\frac{\mathrm{2}{c}\left(−{b}\:\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}{{b}^{\mathrm{2}} −\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)} \\ $$$$\:\:=\frac{\mathrm{2}{c}\left(−{b}\:\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}{\mathrm{4}{ac}} \\ $$$$\:\:=\frac{−{b}\:\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\:\:=\frac{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\therefore\:{Your}\:{formula}\:{is}\:{equivalent}\:{to} \\ $$$${the}\:{quadratic}\:{formula}^{\bullet} \\ $$$$\mathcal{B}{ut}\:{the}\:{quadratic}\:{formula}^{\bullet} \:{is}\:\:{better} \\ $$$${simplified}\:{form}\:{because}\:{it}\:{doesn}'{t} \\ $$$${contain}\:{radical}\:{in}\:{denominator}. \\ $$$$\:\:^{\bullet} \mathcal{H}{ere}\:'{the}\:{quadratic}\:{formula}'\:{means}: \\ $$$$\:\:\:\:\:\:\:{x}=\frac{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$

Commented by mr W last updated on 14/May/20

for the same reason why you write ((√2)/2), not (1/(√2)), and 2+(√3), not (1/(2−(√3))).

$${for}\:{the}\:{same}\:{reason}\:{why}\:{you}\:{write} \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:{not}\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}},\:{and}\:\mathrm{2}+\sqrt{\mathrm{3}},\:{not}\:\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}. \\ $$

Commented by Rio Michael last updated on 14/May/20

okay sirs thank you, was just wondering why this formula is so negleted

$$\mathrm{okay}\:\mathrm{sirs}\:\mathrm{thank}\:\mathrm{you},\:\mathrm{was}\:\mathrm{just}\:\mathrm{wondering}\:\mathrm{why} \\ $$$$\mathrm{this}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{so}\:\mathrm{negleted} \\ $$

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