Question

Solve for $x:4{(x-\frac{1}{x})}^2+8(x+\frac{1}{x})=29,x\ne 0.$

Answer 1

$\begin{array}{c}4{\left(x-\frac{1}{x}\right)}^2+8\left(x+\frac{1}{x}\right)=29\\ 4\left(x^2+\frac{1}{x^2}-2\right)+8x+\frac{8}{x}=29\\ 4x^2+\frac{4}{x^2}-8+8x+\frac{8}{x}=29\\ \frac{4\left(x^4+2x^3-2x^2+2x+1\right)}{x^2}=29\\ Therefore,\\ x=\frac{1}{2},2,-\frac{9}{4}-\frac{\sqrt{65}}{4},\frac{\sqrt{65}}{4}-\frac{9}{4}\end{array}$

Hence, the solution is $x=\frac{1}{2},2,-\frac{9}{4}-\frac{\sqrt{65}}{4},\frac{\sqrt{65}}{4}-\frac{9}{4}$