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Question Number 178785 by Mastermind last updated on 21/Oct/22

Answered by Rasheed.Sindhi last updated on 21/Oct/22

$$$$$$\sqrt[4]{\frac{3-x}{3+x}}-\sqrt{\frac{3+x}{3-x}}=\sqrt{2}$$$$Let\sqrt[4]{\frac{3-x}{3+x}}=y$$$$\sqrt{\frac{3-x}{3+x}}=y^2\Rightarrow \sqrt{\frac{3+x}{3-x}}=\frac{1}{y^2}$$$$y-\frac{1}{y^2}=\sqrt{2}$$$$y^3-\sqrt{2}{y}^2-1=0$$$$\mathcal{T}hisI{can}^{\prime }tsolve.$$$$....$$

Commented by cortano1 last updated on 21/Oct/22

$$\mathrm{typo}.{\mathrm{it}}^{\prime }\mathrm{s}\sqrt[4]{\frac{3-\mathrm{x}}{3+\mathrm{x}}}$$

Commented by Rasheed.Sindhi last updated on 21/Oct/22

$$\mathcal{T}han\mathcal{X}sir,corrected!$$

Commented by mr W last updated on 21/Oct/22

$$y^3-\sqrt{2}{y}^2-1=0$$$$\frac{1}{y^3}+\frac{\sqrt{2}}{y}-1=0$$$$\frac{1}{y}=\sqrt[3]{\sqrt{\frac{1}{4}+\frac{2\sqrt{2}}{27}}+\frac{1}{2}}-\sqrt[3]{\sqrt{\frac{1}{4}+\frac{2\sqrt{2}}{27}}-\frac{1}{2}}$$$$\Rightarrow y=\frac{1}{\sqrt[3]{\sqrt{\frac{1}{4}+\frac{2\sqrt{2}}{27}}+\frac{1}{2}}-\sqrt[3]{\sqrt{\frac{1}{4}+\frac{2\sqrt{2}}{27}}-\frac{1}{2}}}$$

Commented by Spillover last updated on 21/Oct/22

$$\mathrm{thank}\mathrm{you}$$

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