Question Number 46427 by hassentimol last updated on 26/Oct/18
$$\mathrm{Give}\:\mathrm{a}\:\mathrm{proof}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}. \\ $$$$ \\ $$$$ \\ $$$$\forall\:{a}\:\in\:\mathbb{R}^{\ast} \:,\:\left({b},{c}\right)\:\in\:\mathbb{R}\:,\:{x}\:\in\:\mathbb{C}\:: \\ $$$$ \\ $$$$ \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c} \\ $$$$=\:{a}\left({x}−\frac{−{b}\:−\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)\left({x}−\frac{−{b}\:+\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{To}\:\mathrm{answer},\:\mathrm{I}\:\mathrm{should}\:\mathrm{not}\:\mathrm{expand}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{equation}\:\mathrm{but}\:\mathrm{I}\:\mathrm{should}\:\mathrm{start}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{one}. \\ $$$$ \\ $$$$\mathrm{How}\:\mathrm{may}\:\mathrm{I}\:\mathrm{do}\:?\:\mathrm{Thank}\:\mathrm{you}. \\ $$$$ \\ $$
Answered by MrW3 last updated on 26/Oct/18
![ax^2 +bx+c =a(x^2 +(b/a)x+(c/a)) =a(x^2 +2(b/(2a))x+(b^2 /(4a^2 ))−(b^2 /(4a^2 ))+(c/a)) =a[(x+(b/(2a)))^2 −(1/(4a^2 ))(b^2 −4ac)] =a[(x+(b/(2a)))^2 −(((√(b^2 −4ac))/(2a)))^2 ] =a[(x+(b/(2a)))+(((√(b^2 −4ac))/(2a)))][(x+(b/(2a)))−(((√(b^2 −4ac))/(2a)))] =a[x−((−b−(√(b^2 −4ac)))/(2a))][x−((−b+(√(b^2 −4ac)))/(2a))]](https://www.tinkutara.com/question/Q46428.png)
$${ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$={a}\left({x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}\right) \\ $$$$={a}\left({x}^{\mathrm{2}} +\mathrm{2}\frac{{b}}{\mathrm{2}{a}}{x}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }+\frac{{c}}{{a}}\right) \\ $$$$={a}\left[\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)\right] \\ $$$$={a}\left[\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)^{\mathrm{2}} \right] \\ $$$$={a}\left[\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)+\left(\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)\right]\left[\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)−\left(\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right)\right] \\ $$$$={a}\left[{x}−\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right]\left[{x}−\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\right] \\ $$
Commented by hassentimol last updated on 26/Oct/18
$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{much}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{answer}},\:\boldsymbol{\mathrm{sir}}\:! \\ $$$$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{been}}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{helpful}}\:! \\ $$