Question Number 7161 by Tawakalitu. last updated on 14/Aug/16
$${If}\:\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\:{prove}\:{that}. \\ $$$${x}\:=\:\frac{\mathrm{2}{c}}{−\:{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} \:−\:\mathrm{4}{ac}}}\: \\ $$
Commented by sou1618 last updated on 14/Aug/16
$$\left(\ast\right)\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$ \\ $$$${if}\:{x}=\mathrm{0} \\ $$$$\:\:\:{c}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{0}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}=\mathrm{0} \\ $$$$ \\ $$$${if}\:{x}\neq\mathrm{0} \\ $$$$\left(\ast\right)\boldsymbol{\div}{x}^{\mathrm{2}} \:\Rightarrow\:{a}+\frac{{b}}{{x}}+\frac{{c}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$${c}\left\{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{{b}}{{c}}\left(\frac{\mathrm{1}}{{x}}\right)+\frac{{a}}{{c}}\right\}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left\{\frac{{b}}{\mathrm{2}{c}}\left(\frac{\mathrm{1}}{{x}}\right)\right\}+\left\{\left(\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} \right\}+\frac{{a}}{{c}}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{4}{c}^{\mathrm{2}} }−\frac{{a}}{{c}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −\mathrm{4}{ac}}{\mathrm{4}{c}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}=\frac{\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$${x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aftab ahmad last updated on 14/Aug/16
$$\boldsymbol{{N}}{ice}\:{approach}\:{sir}.\: \\ $$
Commented by Tawakalitu. last updated on 14/Aug/16
$${Thanks}\:{so}\:{much}\:{sir}.\:{i}\:{really}\:{appreciate}. \\ $$
Answered by Yozzia last updated on 14/Aug/16
$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\:\:\left({a}\neq\mathrm{0}\right) \\ $$$$\Rightarrow{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${x}=\frac{\left(−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)} \\ $$$${x}=\frac{{b}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{4}{ac}}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)} \\ $$$${x}=\frac{\mathrm{4}{ac}}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}=\frac{\mathrm{2}{c}}{−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$$${This}\:{is}\:{equivalent}\:{to}\: \\ $$$${x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}. \\ $$$$ \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 14/Aug/16
$${Thank}\:{you}\:{very}\:{much}\:{sir}\:{for}\:{your}\:{effort}. \\ $$