Give a proof for the following. ∀ a ∈ R^∗ , (b,c) ∈ R , x ∈ C : ax^2 + bx + c = a(x−((−b − (√(b^2 −4ac)))/(2a)))(x−((−b + (√(b^2 −4ac)))/(2a))) To answer, I should not expand from the second equation but I should start from the first one. How may I do ? Thank you.


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Question Number 46427 by hassentimol last updated on 26/Oct/18

$Give a proof for the following .$ $\forall a \in R^{*}, (b, c) \in R, x \in C :$ ${a x}^{2} + b x + c$ $= a (x - \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}) (x - \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a})$ $To answer, I should not expand from the$ $second equation but I should start from the$ $first one .$ $How may I do ? Thank you .$
	Answered by MrW3 last updated on 26/Oct/18

	${a x}^{2} + b x + c$ $= a (x^{2} + \frac{b}{a} x + \frac{c}{a})$ $= a (x^{2} + 2 \frac{b}{2 a} x + \frac{b^{2}}{4 a^{2}} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a})$ $= a [{(x + \frac{b}{2 a})}^{2} - \frac{1}{4 a^{2}} (b^{2} - 4 a c)]$ $= a [{(x + \frac{b}{2 a})}^{2} - {(\frac{\sqrt{b^{2} - 4 a c}}{2 a})}^{2}]$ $= a [(x + \frac{b}{2 a}) + (\frac{\sqrt{b^{2} - 4 a c}}{2 a})] [(x + \frac{b}{2 a}) - (\frac{\sqrt{b^{2} - 4 a c}}{2 a})]$ $= a [x - \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}] [x - \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}]$
	Commented by hassentimol last updated on 26/Oct/18

	$Thank you very much for your answer, sir!$ $It has been very helpful!$
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