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

$$If{ax}^2+bx+c=0,provethat.$$$$x=\frac{2c}{-b\pm \sqrt{b^2-4ac}}$$

Commented by sou1618 last updated on 14/Aug/16

$$(\ast ){ax}^2+bx+c=0$$$$$$$$ifx=0$$$$c=0\Rightarrow x=\frac{0}{-b\pm \sqrt{b^2-4ac}}=0$$$$$$$$ifx\ne 0$$$$(\ast )\mathbf{÷}{x}^2\Rightarrow a+\frac{b}{x}+\frac{c}{x^2}=0$$$$c\{{\left(\frac{1}{x}\right)}^2+\frac{b}{c}\left(\frac{1}{x}\right)+\frac{a}{c}\}=0$$$${\left(\frac{1}{x}\right)}^2+2\left\{\frac{b}{2c}\left(\frac{1}{x}\right)\right\}+\{{\left(\frac{b}{2c}\right)}^2-{\left(\frac{b}{2c}\right)}^2\}+\frac{a}{c}=0$$$${(\frac{1}{x}+\frac{b}{2c})}^2=\frac{b^2}{4c^2}-\frac{a}{c}$$$${(\frac{1}{x}+\frac{b}{2c})}^2=\frac{b^2-4ac}{4c^2}$$$$\frac{1}{x}+\frac{b}{2c}=\frac{\pm \sqrt{b^2-4ac}}{2c}$$$$\frac{1}{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2c}$$$$x=\frac{2c}{-b\pm \sqrt{b^2-4ac}}$$$$$$$$$$$$$$$$$$

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

$$\mathit{N}iceapproachsir.$$

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

$$Thankssomuchsir.ireallyappreciate.$$

Answered by Yozzia last updated on 14/Aug/16

$${ax}^2+bx+c=0(a\ne 0)$$$$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$$$x=\frac{(-b\pm \sqrt{b^2-4ac})(-b\mp \sqrt{b^2-4ac})}{2a(-b\mp \sqrt{b^2-4ac})}$$$$x=\frac{b^2-b^2+4ac}{2a(-b\mp \sqrt{b^2-4ac})}$$$$x=\frac{4ac}{2a(-b\mp \sqrt{b^2-4ac})}=\frac{2c}{-b\mp \sqrt{b^2-4ac}}$$$$Thisisequivalentto$$$$x=\frac{2c}{-b\pm \sqrt{b^2-4ac}}.$$$$$$$$$$

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

$$Thankyouverymuchsirforyoureffort.$$

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