Q. If the roots of equation $x^{3} - 12x^{2} + 39x - 28 = 0$ are in AP, then their common difference will be
A) $\pm \ 1$
B) $\pm \ 2$
C) $\pm \ 3$
D) $\pm \ 4$
Explanation
(c)
Since, the given equation is cubic, therefore we take three roots.
Let the roots be $a - b,\ a,\ a + d.$
Sum of three numbers in AP $= 3a = 12$
$\Rightarrow \ a = 4$ is a root.
$\therefore\ \$The given equation $x^{3} - 12x^{2} + 39x - 28 = 0$ can be rewritten as
$(x - 4)\left( x^{2} - 8x + 7 \right) = 0$
$\therefore\ \ x = 1,\ 4,\ 7\ \ or\ 7,\ 4,\ 1$
$\therefore\ \ d = \pm 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.