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