Q. The roots of the equation $x^{4} - 2x^{3} + x = 380$ are
A) $5,\ - 4,\ \frac{1 \pm 5\sqrt{- 3}}{2}$
B) $- 5,\ 4,\ \frac{- 1 \pm 5\sqrt{- 3}}{2}$
C) $5,\ 4,\ \frac{- 1 \pm 5\sqrt{- 3}}{2}$
D) $- 5,\ - 4,\ \frac{1 \pm 5\sqrt{3}}{2}$
Explanation
(a)
Given equation is $x^{4} - 2x^{3} + x - 380 = 0$
$\Rightarrow \ \ \ (x - 5)(x + 4)\left( x^{2} - 1 + 19 \right) = 0$
Now, roots of $x^{2} - x + 19$ are
$\frac{1 \pm \sqrt{1 - 4 \times 19}}{2} = \frac{1 \pm 5\sqrt{- 3}}{2}$
$\therefore$ Roots are $5,\ - 4,\ \frac{1 + 5\sqrt{- 3}}{2},\ \frac{1 - 5\sqrt{- 3}}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.