If-ax-2-bx-c-0-prove-that-x-2c-b-b-2-4ac-

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

I f {a x}^{2} + b x + c = 0, p r o v e t h a t .

x = \frac{2 c}{- b \pm \sqrt{b^{2} - 4 a c}}

Commented by sou1618 last updated on 14/Aug/16

(*) {a x}^{2} + b x + c = 0

i f x = 0

c = 0 \Rightarrow x = \frac{0}{- b \pm \sqrt{b^{2} - 4 a c}} = 0

i f x \neq 0

(*) \boldsymbol \div x^{2} \Rightarrow a + \frac{b}{x} + \frac{c}{x^{2}} = 0

c {{(\frac{1}{x})}^{2} + \frac{b}{c} (\frac{1}{x}) + \frac{a}{c}} = 0

{(\frac{1}{x})}^{2} + 2 {\frac{b}{2 c} (\frac{1}{x})} + {{(\frac{b}{2 c})}^{2} - {(\frac{b}{2 c})}^{2}} + \frac{a}{c} = 0

{(\frac{1}{x} + \frac{b}{2 c})}^{2} = \frac{b^{2}}{4 c^{2}} - \frac{a}{c}

{(\frac{1}{x} + \frac{b}{2 c})}^{2} = \frac{b^{2} - 4 a c}{4 c^{2}}

\frac{1}{x} + \frac{b}{2 c} = \frac{\pm \sqrt{b^{2} - 4 a c}}{2 c}

\frac{1}{x} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 c}

x = \frac{2 c}{- b \pm \sqrt{b^{2} - 4 a c}}

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

\boldsymbol N i c e a p p r o a c h s i r .

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

T h a n k s s o m u c h s i r . i r e a l l y a p p r e c i a t e .

Answered by Yozzia last updated on 14/Aug/16

{a x}^{2} + b x + c = 0 (a \neq 0)

\Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{(- b \pm \sqrt{b^{2} - 4 a c}) (- b \mp \sqrt{b^{2} - 4 a c})}{2 a (- b \mp \sqrt{b^{2} - 4 a c})}

x = \frac{b^{2} - b^{2} + 4 a c}{2 a (- b \mp \sqrt{b^{2} - 4 a c})}

x = \frac{4 a c}{2 a (- b \mp \sqrt{b^{2} - 4 a c})} = \frac{2 c}{- b \mp \sqrt{b^{2} - 4 a c}}

T h i s i s e q u i v a l e n t t o

x = \frac{2 c}{- b \pm \sqrt{b^{2} - 4 a c}} .

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

T h a n k y o u v e r y m u c h s i r f o r y o u r e f f o r t .

T h a n k y o u v e r y m u c h s i r f o r y o u r e f f o r t .