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Question-157797




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=(√((4/(4−x^2 ))−1))  squaring and 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∨x^2 =2±(√3)∨x^2 =2±(√2)  testing all solutions ⇒  x^2 =2−(√3)∨x^2 =2+(√2)∨x^2 =2+(√3)  ⇒  x=±(((√6)−(√2))/2)∨x=±(√(2+(√2)))∨x=±(((√6)+(√2))/2)
$${x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}=\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{x}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${x}^{\mathrm{10}} −\mathrm{10}{x}^{\mathrm{8}} +\mathrm{35}{x}^{\mathrm{6}} −\mathrm{50}{x}^{\mathrm{4}} +\mathrm{26}{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}\right)\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}\vee{x}^{\mathrm{2}} =\mathrm{2}\pm\sqrt{\mathrm{3}}\vee{x}^{\mathrm{2}} =\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{solutions}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{3}}\vee{x}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{2}}\vee{x}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${x}=\pm\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\vee{x}=\pm\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\vee{x}=\pm\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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