Question

Solve the equations:
$4x^6-24x^5+57x^4-73x^2-24x+4=0$.

Answer 1

Divide all through by $x^3$ and rearrange;-

$4(x^3+\frac{1}{x^3})-24(x^2+\frac{1}{x^2})+57(x+\frac{1}{x})-73=0$

$4{(x+\frac{1}{x})}^3-24{(x+\frac{1}{x})}^2+57(x+\frac{1}{x})-25=0$

On solving we get the root as;-

$1,\frac{5}{2},\frac{5}{2}$

$x+\frac{1}{x}=1,x+\frac{1}{x}=\frac{5}{2},x+\frac{1}{x}=\frac{5}{2}$

$x+\frac{1}{x}=1$ gives roots $w,w^2$

$x+\frac{1}{x}=\frac{5}{2}$ gives roots $2,\frac{1}{2}$