Find other root of equation $4x^2+ 2x –1 = 0$. If $k$ be a root of the equation $4x^2+ 2x -1 = 0$ then prove that $4k^3-3k$ is the other root.

I simply get the other root as $-\dfrac{1}{4k}$.

As product of roots is $-\dfrac{1}{4}$.

But how to prove that?
Let the other root be given by $\alpha $. Then we have, $$ k+\alpha =- 0.5 \text {  and  } k\alpha =-0.25$$ Then, $k-\alpha = \sqrt { (k+\alpha)^2-4k\alpha} = \sqrt {0.25+1} =\frac {\sqrt {5}}{2} $

Thus, $k = \frac {\sqrt {5}-1}{4} $ and $\alpha = -\frac {\sqrt {5}+1}{4} $. We know that $\sin 18^\circ = \frac {\sqrt {5}-1}{4} = k $ and $\sin 54^\circ = \frac {\sqrt {5}+1}{4} = -\alpha = \sin (3\times 18^\circ) = -4\sin^3 18^\circ + 3\sin 18^\circ = -4k^3+3k $ giving us $\alpha =4k^3-3k $ proving the result. Hope it helps.
Consider $2x=T$.
You can then solve it quickly.