If α is a root of the equation  $4x^2+2x-1=0$ then the other root is given by 

$4x^2+2x-1=0$
α is a root $\Rightarrow 4{\alpha }^2+2\alpha -1=0$
Now  $4{\alpha }^3-3\alpha =\left(\alpha -\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)(4{\alpha }^2+2\alpha -1)-\alpha -\frac{1}{2}$ $=-\alpha -\frac{1}{2}$
If ß is a root then $\alpha +\beta =-\frac{1}{2},\beta =-\frac{1}{2}-\alpha$
Hence $4{\alpha }^3-3\alpha$ is other root
  (or) $\alpha =\frac{\sqrt{5}-1}{4}=\sin {18}^{°}$
$\beta =-\left(\frac{{\displaystyle \sqrt{5}+1}}{{\displaystyle 4}}\right)=-\sin {54}^{°}$