Q. The roots of $|x - 2|^{2} + |x - 2| - 6 = 0$ are
A) 4, 2
B) 0, 4
C) -1, 3
D) 5, 1
Explanation
(b)
Let $|x - 2| = y$
$\therefore\ \ \ y^{2} + y - 6 = 0$
$\Rightarrow \ \ y = - 3,\ 2$
$\Rightarrow \ \ \ |x - 2| = - 3,\ \ \ |x - 2| = 2$
$\Rightarrow \ \ \pm (x - 2) = 2$ [$\because|x - 2|$ cannot be negative]
$\therefore\ \ \ x = 4,\ 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.