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Question Number 99319 by  M±th+et+s last updated on 20/Jun/20

$$solve$$$$(x^2-\frac{1}{x^2})-11(x-\frac{1}{x})=14$$

Answered by behi83417@gmail.com last updated on 20/Jun/20

$$\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow {(\mathrm{x}+\frac{1}{\mathrm{x}})}^2={(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+2={\mathrm{t}}^2+2$$$$\mathrm{t}\sqrt{{\mathrm{t}}^2+2}-11\mathrm{t}=14\Rightarrow \mathrm{t}\sqrt{{\mathrm{t}}^2+2}=11\mathrm{t}+14$$$$\Rightarrow {\mathrm{t}}^2({\mathrm{t}}^2+2)={121\mathrm{t}}^2+308\mathrm{t}+196$$$$\Rightarrow {\mathrm{t}}^4-{119\mathrm{t}}^2-308\mathrm{t}-196=0$$$$\Rightarrow \mathrm{t}=[-9.41,-1.57,-1.1,12.077]$$$$\{\begin{array}{l}\mathrm{x}-\frac{1}{\mathrm{x}}=-9.41\Rightarrow {\mathrm{x}}^2+9.41\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{-9.41\pm \sqrt{9.{41}^2+4}}{2}=0.11,-9.5\end{array}$$$$\{\begin{array}{l}\mathrm{x}-\frac{1}{\mathrm{x}}=-1.57\Rightarrow {\mathrm{x}}^2+1.57\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{-1.57\pm \sqrt{1.{57}^2+4}}{2}=0.49,-2.1\end{array}$$$$\{\begin{array}{l}\mathrm{x}-\frac{1}{\mathrm{x}}=-1.1\Rightarrow {\mathrm{x}}^2+1.1\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{-1.1\pm \sqrt{1.1^2+4}}{2}=0.6,-1.7\end{array}$$$$\{\begin{array}{l}\mathrm{x}-\frac{1}{\mathrm{x}}=12.1\Rightarrow {\mathrm{x}}^2-12.1\mathrm{x}-1=0\\ \Rightarrow \mathrm{x}=\frac{12.1\pm \sqrt{12.1^2+4}}{2}=12.18,-0.082\end{array}$$

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