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Question Number 7161 by Tawakalitu. last updated on 14/Aug/16

If ax^2 + bx + c = 0, prove that. x = ((2c)/(− b ± (√(b^2 − 4ac))))

$${If}\:\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\:{prove}\:{that}. \\ $$$${x}\:=\:\frac{\mathrm{2}{c}}{−\:{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} \:−\:\mathrm{4}{ac}}}\: \\ $$

Commented by sou1618 last updated on 14/Aug/16

(∗) ax^2 +bx+c=0 if x=0 c=0⇒x=(0/(−b±(√(b^2 −4ac))))=0 if x≠0 (∗)÷x^2 ⇒ a+(b/x)+(c/x^2 )=0 c{((1/x))^2 +(b/c)((1/x))+(a/c)}=0 ((1/x))^2 +2{(b/(2c))((1/x))}+{((b/(2c)))^2 −((b/(2c)))^2 }+(a/c)=0 ((1/x)+(b/(2c)))^2 =(b^2 /(4c^2 ))−(a/c) ((1/x)+(b/(2c)))^2 =((b^2 −4ac)/(4c^2 )) (1/x)+(b/(2c))=((±(√(b^2 −4ac)))/(2c)) (1/x)=((−b±(√(b^2 −4ac)))/(2c)) x=((2c)/(−b±(√(b^2 −4ac))))

$$\left(\ast\right)\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$ \\ $$$${if}\:{x}=\mathrm{0} \\ $$$$\:\:\:{c}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{0}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}=\mathrm{0} \\ $$$$ \\ $$$${if}\:{x}\neq\mathrm{0} \\ $$$$\left(\ast\right)\boldsymbol{\div}{x}^{\mathrm{2}} \:\Rightarrow\:{a}+\frac{{b}}{{x}}+\frac{{c}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$${c}\left\{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{{b}}{{c}}\left(\frac{\mathrm{1}}{{x}}\right)+\frac{{a}}{{c}}\right\}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left\{\frac{{b}}{\mathrm{2}{c}}\left(\frac{\mathrm{1}}{{x}}\right)\right\}+\left\{\left(\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} \right\}+\frac{{a}}{{c}}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{4}{c}^{\mathrm{2}} }−\frac{{a}}{{c}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}\right)^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} −\mathrm{4}{ac}}{\mathrm{4}{c}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{{b}}{\mathrm{2}{c}}=\frac{\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$${x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by aftab ahmad last updated on 14/Aug/16

Nice approach sir.

$$\boldsymbol{{N}}{ice}\:{approach}\:{sir}.\: \\ $$

Commented by Tawakalitu. last updated on 14/Aug/16

Thanks so much sir. i really appreciate.

$${Thanks}\:{so}\:{much}\:{sir}.\:{i}\:{really}\:{appreciate}. \\ $$

Answered by Yozzia last updated on 14/Aug/16

ax^2 +bx+c=0 (a≠0) ⇒x=((−b±(√(b^2 −4ac)))/(2a)) x=(((−b±(√(b^2 −4ac)))(−b∓(√(b^2 −4ac))))/(2a(−b∓(√(b^2 −4ac))))) x=((b^2 −b^2 +4ac)/(2a(−b∓(√(b^2 −4ac))))) x=((4ac)/(2a(−b∓(√(b^2 −4ac)))))=((2c)/(−b∓(√(b^2 −4ac)))) This is equivalent to x=((2c)/(−b±(√(b^2 −4ac)))).

$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\:\:\left({a}\neq\mathrm{0}\right) \\ $$$$\Rightarrow{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${x}=\frac{\left(−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)} \\ $$$${x}=\frac{{b}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{4}{ac}}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)} \\ $$$${x}=\frac{\mathrm{4}{ac}}{\mathrm{2}{a}\left(−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)}=\frac{\mathrm{2}{c}}{−{b}\mp\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$$${This}\:{is}\:{equivalent}\:{to}\: \\ $$$${x}=\frac{\mathrm{2}{c}}{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}. \\ $$$$ \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 14/Aug/16

Thank you very much sir for your effort.

$${Thank}\:{you}\:{very}\:{much}\:{sir}\:{for}\:{your}\:{effort}. \\ $$

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