# Proving the second root of a quadratic equation

If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expressions of $\alpha$ and check whether manipulating the first root gives me the second root (which seems much complicated), or is there any other way?

• Check $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1 = 0$ and $4\alpha^3-3\alpha \ne \alpha$. – njguliyev Nov 3 '13 at 12:03

Set $x=4\alpha^2-3\alpha$ in the original polynomial equation, and check to see that the left side reduces to zero under the assumption that $4\alpha^2+2\alpha-1=0$, or equivalently that $\alpha^2 = \frac 14(-2\alpha+1)$.

The other root $\beta$ is determined by the properties $\alpha+\beta=-\frac 12$ and $\alpha\beta=-\frac14$. Use polynomial division to show that $(4\alpha^3-3\alpha)+\alpha = [???]\cdot(4\alpha^2+2\alpha-1)-\frac12$ and $(4\alpha^3-3\alpha)\cdot\alpha = [???]\cdot(4\alpha^2+2\alpha-1)-\frac14$

Clearly, the absolute value of both roots are $<1$

$$0=(4x^2+2x-1)(x-1)=4x^3-3x-(2x^2-1)$$

If $x=\cos y,$ we have $$\cos3y=\cos2y$$

$\implies3y=360^\circ n\pm2y$ where $n$ is any integer.

'+'$\implies y=360^\circ n\implies x=1$

'-'$\implies y=72^\circ n$ where $n\equiv0,\pm1,\pm2\pmod5$

As $x=1$ does not satisfy $4x^2+2x-1=0,n\not\equiv0\pmod5$

So, the roots of $4x^2+2x-1=0$ are $\cos72^\circ,\cos144^\circ=-\cos36^\circ$

Now $4\cos^372^\circ-3\cos72^\circ=\cos3\cdot72^\circ=\cos216^\circ=\cdots=-\cos36^\circ$ ?

Similarly, $\cos3\cdot(144^\circ)=\cdots=\cos72^\circ$

Can you take it from here