Question

Solve the equation $4\left(x^2+\frac{1}{x^2}\right)+8\left(x-\frac{1}{x}\right)-29=0$

Answer 1

$4(x^2+\frac{1}{x^2})+8(x+\frac{1}{x})-29=0$

$4[{(x-\frac{1}{x})}^2+2]+8(x-\frac{1}{x})-29=0$

Let $x-\frac{1}{x}=t$

$4t^2+8t-21=0$

$4t^2+14t-6t-21=0$

$2+(2t+7)-3(2t+7)=0$

$(2t-3)(2t+7)=0$

$t=\frac{3}{2},\frac{-7}{2}$

$x-\frac{1}{x}=\frac{3}{2}$, and $x-\frac{1}{x}=\frac{-7}{2}$

$2x^2-3x-2=0$ $2x^2+7x-2=0$

$2x^2-4x+x-2=0$ $x=\frac{-7\pm \sqrt{49+16}}{4}$

$(2x+1)(x-2)=0$ $=\frac{-7\pm \sqrt{65}}{4}$

$x=\frac{-1}{2},2$.