Solve-for-x-in-the-equation-below-ax-2-bx-c-0-

Question Number 119376 by Abdulmajeed last updated on 24/Oct/20

S o l v e f o r x i n t h e e q u a t i o n b e l o w

{a x}^{2} + b x + c = 0 .

Commented by ZiYangLee last updated on 24/Oct/20

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

Commented by peter frank last updated on 24/Oct/20

use complete the square

Answered by peter frank last updated on 24/Oct/20

x^{2} + \frac{b}{a} x = - \frac{c}{a}

add {(\frac{b}{2 a})}^{2} both sides

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

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

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

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

- \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 ac}}{2 a}

x = \frac{- b + \sqrt{b^{2} - 4 ac}}{2 a}

Commented by ZiYangLee last updated on 24/Oct/20

Nice

Answered by TANMAY PANACEA last updated on 24/Oct/20

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

{(2 a x)}^{2} + 2 \times 2 a x \times b + b^{2} + 4 a c - b^{2} = 0

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

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

Answered by malwan last updated on 25/Oct/20

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

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

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

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

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

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

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