Question

The roots of the equation $x^4-2x^3+x=380$ are

• A. $5,-4,\raisebox{1ex}{$\left[1\pm 5\sqrt{-3}\right]$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $-5,4,\raisebox{1ex}{$\left[-1\pm 5\sqrt{-3}\right]$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $5,4,\raisebox{1ex}{$\left[-1\pm 5\sqrt{-3}\right]$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• D. $-5,-4,\raisebox{1ex}{$\left[1\pm 5\sqrt{3}\right]$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

The correct option is A

$5,-4,\raisebox{1ex}{$\left[1\pm 5\sqrt{-3}\right]$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Explanation for the correct option

The given equation is:

$x^4-2x^3+x=380$

$$\begin{gathered} \Rightarrow (x^2-x-20)(x^2-x+19)=0 \\ \Rightarrow (x-5)(x+4)(x^2-x+19)=0 \\ \Rightarrow x-5=0,x+4=0,x^2-x+19=0 \end{gathered}$$

$\Rightarrow x=5,x=-4$

For $x^2-x+19=0$ roots of x is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

According to $x^2-x+19=0$ the value of $b=-1,a=1$ and $c=19$

$$\begin{gathered} x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\left(1\right)\left(19\right)}}{2\left(1\right)} \\ x=\frac{1\pm \sqrt{1-76}}{2} \\ x=\frac{1\pm \sqrt{-75}}{2} \\ x=\frac{1\pm 5\sqrt{3}i}{2}[\because b^2-4ac<0] \\ x=\frac{1\pm 5\sqrt{-3}}{2} \end{gathered}$$

The roots of the equation are $5,-4,\frac{\left[1\pm 5\sqrt{-3}\right]}{2}$

Hence the correct option is option A