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

$$Solvefor\mathit{x}intheequationbelow$$$${ax}^2+bx+c=0.$$

Commented by ZiYangLee last updated on 24/Oct/20

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

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

$$\mathrm{use}\mathrm{complete}\mathrm{the}\mathrm{square}$$

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

$${\mathrm{x}}^2+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}=-\frac{\mathrm{c}}{\mathrm{a}}$$$$\mathrm{add}{\left(\frac{\mathrm{b}}{2\mathrm{a}}\right)}^2\mathrm{both}\mathrm{sides}$$$${\mathrm{x}}^2+\frac{\mathrm{b}}{\mathrm{a}}\mathrm{x}+{\left(\frac{\mathrm{b}}{2\mathrm{a}}\right)}^2=-\frac{\mathrm{c}}{\mathrm{a}}+\frac{{\mathrm{b}}^2}{{4\mathrm{a}}^2}$$$${(\mathrm{x}+\frac{\mathrm{b}}{2\mathrm{a}})}^2=\frac{{\mathrm{b}}^2-4\mathrm{ac}}{{4\mathrm{a}}^2}$$$$\mathrm{x}+\frac{\mathrm{b}}{2\mathrm{a}}=\pm \sqrt{\frac{{\mathrm{b}}^2-4\mathrm{ac}}{{4\mathrm{a}}^2}}$$$$\mathrm{x}=-\frac{\mathrm{b}}{2\mathrm{a}}\pm \sqrt{\frac{{\mathrm{b}}^2-4\mathrm{ac}}{{4\mathrm{a}}^2}}$$$$-\frac{\mathrm{b}}{2\mathrm{a}}\pm \frac{\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$$$$\mathrm{x}=\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$$

Commented by ZiYangLee last updated on 24/Oct/20

$$\mathrm{Nice}$$

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

$$4a({ax}^2+bx+c)=0$$$${\left(2ax\right)}^2+2\times 2ax\times b+b^2+4ac-b^2=0$$$${(2ax+b)}^2=b^2-4ac$$$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Answered by malwan last updated on 25/Oct/20

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

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