Question
Solve $\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$
✓
Answer
$\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$
Let $x-\frac{1}{x}=\mathrm{y}$
squaring on both sides
squaring on both sides
$ \begin{aligned} & \left(x-\frac{1}{x}\right)^2=\mathrm{y}^2 \\ & x^2+\frac{1}{x^2}-2=y^2 \\ & \Rightarrow x^2+\frac{1}{x^2}=y^2+2 \\ & \mathrm{~m}^2+2-3(\mathrm{~m})-2=0 \\ & \mathrm{~m}^2-3 \mathrm{~m}=0 \\ & \mathrm{~m}(\mathrm{~m}-3)=0 \\ & \mathrm{~m}=0 \text { and } \mathrm{m}-3=0 \\ & x-\frac{1}{x}=0 \text { or } x-\frac{1}{x}=3 \\ & \frac{x^2-1}{x}=0 \text { or } \frac{x^2-1}{x}=3 \\ & =\mathrm{x}^2-1=0 \text { or } \mathrm{x}^2-1=3 \mathrm{x} \\ & \mathrm{x}^2=1 \text { or } \mathrm{x}^2-3 \mathrm{x}-1=0 \end{aligned} $
$\begin{aligned} & x=\sqrt{1} \text { or } x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)} \\ & x= \pm 1 \text { or }=x=\frac{3 \pm \sqrt{13}}{2}\end{aligned}$
Let $x-\frac{1}{x}=\mathrm{y}$
squaring on both sides
squaring on both sides
$ \begin{aligned} & \left(x-\frac{1}{x}\right)^2=\mathrm{y}^2 \\ & x^2+\frac{1}{x^2}-2=y^2 \\ & \Rightarrow x^2+\frac{1}{x^2}=y^2+2 \\ & \mathrm{~m}^2+2-3(\mathrm{~m})-2=0 \\ & \mathrm{~m}^2-3 \mathrm{~m}=0 \\ & \mathrm{~m}(\mathrm{~m}-3)=0 \\ & \mathrm{~m}=0 \text { and } \mathrm{m}-3=0 \\ & x-\frac{1}{x}=0 \text { or } x-\frac{1}{x}=3 \\ & \frac{x^2-1}{x}=0 \text { or } \frac{x^2-1}{x}=3 \\ & =\mathrm{x}^2-1=0 \text { or } \mathrm{x}^2-1=3 \mathrm{x} \\ & \mathrm{x}^2=1 \text { or } \mathrm{x}^2-3 \mathrm{x}-1=0 \end{aligned} $
$\begin{aligned} & x=\sqrt{1} \text { or } x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)} \\ & x= \pm 1 \text { or }=x=\frac{3 \pm \sqrt{13}}{2}\end{aligned}$
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