Problem of quadratic equation If $\alpha$ be a root of $ 4x^2 +2x -1 = 0 $ , prove that the other root is $4\alpha^3  - 3\alpha$ . I have tried to do it but of no success.[$4\alpha^3 -2\alpha$ = $\dfrac {-1}{2}$  and $4\alpha^4 - 3\alpha^2$ = $\dfrac {-1}{4}$ ] .How to prove it?
 A: $\, 0=(4\alpha^2\!-\!1)^2\!-\!(2\alpha)^2\! =\, \color{#0a0}4\alpha(4\alpha^3\!-3\alpha)\!+\!\color{#c00}1\,\Rightarrow\, \alpha(4\alpha^3\!-\!3\alpha) \,=\, {-}\dfrac{\color{#c00}1}{\color{#0a0}4}\,=\, $ product of roots
A: Someone will probably show you a smart way to do this, but ...I guess you could just use that $\alpha^2 = 1 - 2\alpha$ and then
$$\begin{align}
4(4\alpha^3 - 3\alpha)^2 + &2(4\alpha^3 - 3\alpha) - 1 \\ &= 4(4\alpha^6 + 9\alpha^2 - 24\alpha^4) + 8\alpha^3 - 6\alpha - 1 \\
&= 16\alpha^6 + 36\alpha^2 - 96\alpha^4 + 8\alpha^3 - 6\alpha -1\\
&= 16(1-2\alpha)^3 + 36(1 - 2\alpha) - 96(1-2\alpha)^2 + 8\alpha(1-2\alpha) - 6\alpha -1 \\
&= \dots
\end{align}
$$
A: Write $4x^3-3x=(4x^2+2x-1)(x-1/2)-(x+1/2)$.
Plug $x=\alpha$ and get $4\alpha^3  - 3\alpha=-\alpha-1/2$, which is the other root because the sum of the roots is $-1/2$.
A: Suppose that this is true.
In that case, if you let $\alpha = \cos\theta$ then the two roots are $\cos(\theta)$ and $\cos(3\theta)$ using the identity $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$.  
Take any one root of the equation and hence deduce $\theta$  using $\theta = \cos^{-1}(\alpha)$.  If $\cos(3\theta)$ evaluates to the other root, then $4\alpha^3 - 3\alpha$ is indeed your other root.
A: If $\alpha$ is a root then
$$
4 \alpha^2 + 2 \alpha - 1 = 0.
$$
Therefore
$$
\begin{eqnarray}
4 \Big( -\tfrac{1}{2} - \alpha \Big)^2 + 2 \Big( -\tfrac{1}{2} - \alpha \Big) - 1 &=&
4 \Big( \tfrac{1}{4} + \alpha + \alpha^2 \Big)
 - 2 \Big( \tfrac{1}{2} + \alpha \Big) - 1\\
&=& 4 \alpha^2 + 2 \alpha - 1\\
&=& 0.
\end{eqnarray}
$$
So if $\alpha$ is a root, then $- \tfrac{1}{2} - \alpha$ is also a root, but
$$
\begin{eqnarray}
4 \alpha^3 - 3 \alpha &=& \alpha \Big( 4\alpha^2 - 3 \Big)\\
&=& \alpha \Big( 1 - 2 \alpha - 3 \Big)\\
&=& - \Big( 2 \alpha^2 + 2 \alpha \Big)\\
&=& - \Big( \frac{1}{2} - \alpha + 2 \alpha \Big)\\
&=& - \frac{1}{2} - \alpha,
\end{eqnarray}
$$
So $ 4 \alpha^3 - 3\alpha$ is also a root.
