Question Number 119376 by Abdulmajeed last updated on 24/Oct/20
$${Solve}\:{for}\:\boldsymbol{{x}}\:{in}\:{the}\:{equation}\:{below} \\ $$$${ax}^{\mathrm{2}} \:+{bx}\:+\:{c}\:=\:\mathrm{0}. \\ $$
Commented by ZiYangLee last updated on 24/Oct/20
$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$
Commented by peter frank last updated on 24/Oct/20
$$\mathrm{use}\:\mathrm{complete}\:\mathrm{the}\:\mathrm{square} \\ $$
Answered by peter frank last updated on 24/Oct/20
$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}=−\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\mathrm{add}\:\:\left(\frac{\mathrm{b}}{\mathrm{2a}}\right)^{\mathrm{2}} \:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}+\left(\frac{\mathrm{b}}{\mathrm{2a}}\right)^{\mathrm{2}} =−\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{4a}^{\mathrm{2}} } \\ $$$$\left(\mathrm{x}+\frac{\mathrm{b}}{\mathrm{2a}}\right)^{\mathrm{2}} =\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}{\mathrm{4a}^{\mathrm{2}} } \\ $$$$\mathrm{x}+\frac{\mathrm{b}}{\mathrm{2a}}=\pm\sqrt{\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}{\mathrm{4a}^{\mathrm{2}} }} \\ $$$$\mathrm{x}=−\frac{\mathrm{b}}{\mathrm{2a}}\pm\sqrt{\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}{\mathrm{4a}^{\mathrm{2}} }} \\ $$$$−\frac{\mathrm{b}}{\mathrm{2a}}\pm\frac{\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\mathrm{x}=\frac{−\mathrm{b}+\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$
Commented by ZiYangLee last updated on 24/Oct/20
$$\mathrm{Nice} \\ $$
Answered by TANMAY PANACEA last updated on 24/Oct/20
$$\mathrm{4}{a}\left({ax}^{\mathrm{2}} +{bx}+{c}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{ax}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{2}{ax}×{b}+{b}^{\mathrm{2}} +\mathrm{4}{ac}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{ax}+{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$
Answered by malwan last updated on 25/Oct/20
![ax^2 + bx + c = 0 (÷ x^2 ) ⇒ c((1/x))^2 + b((1/x)) + a = 0 c[((1/x))^2 + (b/c)((1/x)) + (a/c) ]= 0 ((1/x))^2 + (b/c)((1/x)) + (b^2 /(4c^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))))](https://www.tinkutara.com/question/Q119626.png)
$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\left(\boldsymbol{\div}\:{x}^{\mathrm{2}} \:\right) \\ $$$$\Rightarrow\:{c}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\:{b}\left(\frac{\mathrm{1}}{{x}}\right)\:+\:{a}\:=\:\mathrm{0} \\ $$$${c}\left[\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\:\frac{{b}}{{c}}\left(\frac{\mathrm{1}}{{x}}\right)\:+\:\frac{{a}}{{c}}\:\right]=\:\mathrm{0} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\:\frac{{b}}{{c}}\left(\frac{\mathrm{1}}{{x}}\right)\:+\:\frac{{b}^{\mathrm{2}} }{\mathrm{4}{c}^{\mathrm{2}} }\:=\:\frac{{b}^{\mathrm{2}} }{\mathrm{4}{c}^{\mathrm{2}} }\:−\:\frac{{a}}{{c}} \\ $$$$\left[\left(\frac{\mathrm{1}}{{x}}\right)\:+\:\frac{{b}}{\mathrm{2}{c}}\:\right]^{\mathrm{2}} \:=\:\frac{{b}^{\mathrm{2}} \:−\:\mathrm{4}{ac}}{\mathrm{4}{c}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{x}}\:+\:\frac{{b}}{\mathrm{2}{c}}\:=\:\pm\:\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$$\frac{\mathrm{1}}{{x}}\:=\:\frac{−{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{c}} \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{2}{c}}{−{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}} \\ $$