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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!