Question

The roots of the equation $x^4+3x^3-3x-1=0$ are:

• A. $-1,1,\frac{-3\pm \sqrt{5}}{2}$
• B. $1,1,\frac{-3\pm \sqrt{5}}{2}$
• C. $-1,-1,\frac{-3\pm \sqrt{5}}{2}$
• D. $-1,1,\frac{3\pm \sqrt{5}}{2}$

Answer 1

Answer: (A)

$-1,1,\frac{-3\pm \sqrt{5}}{2}$

Let $\alpha ,\beta ,\gamma ,\delta$ be the roots of $x^4+3x^3-3x-1=0$
$S_1=\alpha +\beta +\gamma +\delta =-3S_4=\alpha \beta \gamma \delta =-1$
And only option (A) satisfies these condition
$-1+1+\frac{-3+\sqrt{5}}{2}+\frac{-3-\sqrt{5}}{2}=-3$
And
$(-1)\left(1\right)\left(\frac{-3+\sqrt{5}}{2}\right)\left(\frac{-3-\sqrt{5}}{2}\right)=-1$