Question

Solve the following equations:
$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}-\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=8x\sqrt{x^2-3x+2}$.

Answer 1

$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}-\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=8x\sqrt{x^2-3x+2}\frac{{(x+\sqrt{x^2-1})}^2-{(x-\sqrt{x^2-1})}^2}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=8x\sqrt{x^2-3x+2}\frac{x^2+x^2-1+2x\sqrt{x^2-1}-x^2-x^2+1+2x\sqrt{x^2-1}}{x^2-x^2+1}=8x\sqrt{x^2-3x+2}4x\sqrt{x^2-1}=8x\sqrt{x^2-3x+2}x\sqrt{x^2-1}=2x\sqrt{x^2-3x+2}=0x\sqrt{x^2-1}-2x\sqrt{x^2-3x+2}=0x(\sqrt{x^2-1}-2\sqrt{x^2-3x+2})=0$

$\Rightarrow x=0$ and $\sqrt{x^2-1}-2\sqrt{x^2-3x+2}=0$

$\sqrt{x^2-1}=2\sqrt{x^2-3x+2}{x}^2-1=4(x^2-3x+2)3x^2-12x+9=03x^2-9x-3x+9=03x(x-3)-3(x-3)=0(3x-3)(x-3)=0\Rightarrow x=1,3$

So the values of $x$ are $0,1$ and $3$