Question
Solve for $x: \left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
✓
Answer
Given equation
$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
Put $x-\frac{1}{x}=y$, squaring $x^2+\frac{1}{x^2}-2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}=y^2+2$
Then given rquation becomes :
$x^2+\frac{1}{x^2}+2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
$\Rightarrow y^2+2+2-\frac{3}{2} y-4=0$
$\Rightarrow y^2+4-\frac{3}{2} y-4=0$
$\Rightarrow 2 y^2-3 y=0$
$\Rightarrow y(2 y-3)=0$
$\Rightarrow y=0$ or $2 y-3=0$
i.e., $y=\frac{3}{2}$
But $x-\frac{1}{x}=y$
Then $x-\frac{1}{x}=0$
$\Rightarrow x 2-1=0$
$\Rightarrow x = \pm 1$
or
$x-\frac{1}{x}=\frac{3}{2}$
$\Rightarrow \frac{x^2-1}{x}=\frac{3}{2}$
$\Rightarrow 2 x^2-2=3 x$
$\Rightarrow 2 x^2-3 x-2=0$
$\Rightarrow 2 x^2-4 x+x-2=0$
$\Rightarrow 2 x(x-2)+1(x-2)=0$
$\Rightarrow(x-2)(2 x+1)=0$
$\Rightarrow x-2=0 \text { or } 2 x+1=0$
$\Rightarrow x=2 \text { or } x=-\frac{1}{2}$
Hence, $x= \pm 1,2,-\frac{1}{2} .$
$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
Put $x-\frac{1}{x}=y$, squaring $x^2+\frac{1}{x^2}-2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}=y^2+2$
Then given rquation becomes :
$x^2+\frac{1}{x^2}+2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
$\Rightarrow y^2+2+2-\frac{3}{2} y-4=0$
$\Rightarrow y^2+4-\frac{3}{2} y-4=0$
$\Rightarrow 2 y^2-3 y=0$
$\Rightarrow y(2 y-3)=0$
$\Rightarrow y=0$ or $2 y-3=0$
i.e., $y=\frac{3}{2}$
But $x-\frac{1}{x}=y$
Then $x-\frac{1}{x}=0$
$\Rightarrow x 2-1=0$
$\Rightarrow x = \pm 1$
or
$x-\frac{1}{x}=\frac{3}{2}$
$\Rightarrow \frac{x^2-1}{x}=\frac{3}{2}$
$\Rightarrow 2 x^2-2=3 x$
$\Rightarrow 2 x^2-3 x-2=0$
$\Rightarrow 2 x^2-4 x+x-2=0$
$\Rightarrow 2 x(x-2)+1(x-2)=0$
$\Rightarrow(x-2)(2 x+1)=0$
$\Rightarrow x-2=0 \text { or } 2 x+1=0$
$\Rightarrow x=2 \text { or } x=-\frac{1}{2}$
Hence, $x= \pm 1,2,-\frac{1}{2} .$
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