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Question Number 35336 by jasno91 last updated on 18/May/18

$$\sqrt{2x^2}+7x+5\sqrt{2}=0$$

Commented by Rasheed.Sindhi last updated on 18/May/18

$$\mathrm{Is}\mathrm{it}\sqrt{2}{x}^2?$$

Answered by ajfour last updated on 18/May/18

$$\sqrt{2}{x}^2+7x+5\sqrt{2}=0$$$$\sqrt{2}{x}^2+5x+2x+5\sqrt{2}=0$$$$x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$$$$(\sqrt{2}x+5)(x+\sqrt{2})=0$$$$\Rightarrow x=\frac{-5}{\sqrt{2}},-\sqrt{2}.$$

Answered by Rio Mike last updated on 18/May/18

$$x=\frac{-7\pm \sqrt{7^2-4\sqrt{2}\times 5\sqrt{2}}}{2\sqrt{2}}$$$$x=\frac{-7\pm \sqrt{49-4\times 2\times 5}}{2\sqrt{2}}$$$$x=\frac{-7\pm \sqrt{9}}{2\sqrt{2}}$$$$x=\frac{-7\pm 3}{2\sqrt{2}}$$$$x=\frac{-7\pm 3\sqrt{2}}{4}$$$$Eitherx=\frac{-7+3\sqrt{2}}{4}orx=\frac{-7-3\sqrt{2}}{4}$$

Commented by ajfour last updated on 18/May/18

$$secondlastlineshouldbe$$$$x=\frac{-7\sqrt{2}\pm 3\sqrt{2}}{4}$$$$Eitherx=\frac{-5}{\sqrt{2}}orx=-\sqrt{2}.$$

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