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Question Number 142971 by 0731619 last updated on 08/Jun/21

Commented by wassel last updated on 11/Jun/21

$${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=x+\frac{1}{x}-2=\frac{1}{2}-2=-\frac{3}{2}=\frac{3}{2}{i}^2$$$$\sqrt{x}-\frac{1}{\sqrt{x}}=\pm i\sqrt{\frac{3}{2}}$$

Answered by Rasheed.Sindhi last updated on 08/Jun/21

$$x+\frac{1}{x}=\frac{1}{2};\sqrt{x}-\frac{1}{\sqrt{x}}=?$$$${\left(\sqrt{x}\right)}^2+\frac{1}{{\left(\sqrt{x}\right)}^2}=\frac{1}{2}$$$${\left(\sqrt{x}\right)}^2+\frac{1}{{\left(\sqrt{x}\right)}^2}-2=\frac{1}{2}-2=\frac{1-4}{2}$$$${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=-\frac{3}{2}=-\frac{6}{4}={(\pm \frac{i\sqrt{6}}{2})}^2$$$$\sqrt{x}-\frac{1}{\sqrt{x}}=\pm \frac{i\sqrt{6}}{2}$$

Commented by 0731619 last updated on 08/Jun/21

$$tanks$$

Answered by Rasheed.Sindhi last updated on 08/Jun/21

$${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2=x+\frac{1}{x}-2=\frac{1}{2}-2=-\frac{3}{2}$$$$=\frac{-6}{4}={(\pm \frac{i\sqrt{6}}{2})}^2$$$$\sqrt{x}-\frac{1}{\sqrt{x}}=\pm \frac{i\sqrt{6}}{2}$$

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