# Thread: Quartic Equation

1. ## Quartic Equation

Prove that $\displaystyle \tan \frac{\pi}{15}$ is a root of the equation.

$\displaystyle t^4 -6 \sqrt{3}t^3 +8t^2 + 2 \sqrt{3} t -1 =0$

Give the other roots in the form $\displaystyle \tan \frac{r \pi}{15}$
I have tired using an expression for $\displaystyle \tan 4 \theta$ and got nowhere

Bobak

2. Originally Posted by bobak
I have tried using an expression for $\displaystyle \tan 4 \theta$ and got nowhere
I think maybe you're confusing π/15 radians with 15 degrees! Try tan(5θ) instead of tan(4θ). If you put the expression for tan(5θ) equal to √3, you'll get a fifth-degree equation one of whose roots is –√3. Divide out by the factor t+√3 and you'll be on the right track.

3. Originally Posted by Opalg
I think maybe you're confusing π/15 radians with 15 degrees! Try tan(5θ) instead of tan(4θ). If you put the expression for tan(5θ) equal to √3, you'll get a fifth-degree equation one of whose roots is –√3. Divide out by the factor t+√3 and you'll be on the right track.
I never thought of using $\displaystyle \tan(5 \theta)$ at all! So the general solution is given by $\displaystyle \theta = \frac{(3n+1) \pi}{15}$ excluding the solution $\displaystyle \theta = \frac{2 \pi}{3}$ ( i.e. r = 3) as this corrospends to $\displaystyle \tan \theta = - \sqrt{3}$

Many Thanks

Bobak