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Question Number 157797 by akolade last updated on 28/Oct/21

Answered by MJS_new last updated on 28/Oct/21

$$x^4-3x^2+1=\sqrt{\frac{4}{4-x^2}-1}$$$$\mathrm{squaring}\mathrm{and}\mathrm{transforming}$$$$x^{10}-10x^8+35x^6-50x^4+26x^2-4=0$$$$(x^2-2)(x^4-4x^2+1)(x^4-4x^2+2)=0$$$$x^2=2\vee x^2=2\pm \sqrt{3}\vee x^2=2\pm \sqrt{2}$$$$\mathrm{testing}\mathrm{all}\mathrm{solutions}\Rightarrow$$$$x^2=2-\sqrt{3}\vee x^2=2+\sqrt{2}\vee x^2=2+\sqrt{3}$$$$\Rightarrow$$$$x=\pm \frac{\sqrt{6}-\sqrt{2}}{2}\vee x=\pm \sqrt{2+\sqrt{2}}\vee x=\pm \frac{\sqrt{6}+\sqrt{2}}{2}$$

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