For $x \in \mathbb{R}$, then number of real roots of the equation $3x^{2} - 4|x^{2} - 1| + x - 1 = 0$ is ____.

$3x^2 - 4|x^2 - 1| + x - 1 = 0$ $x \in \mathbb{R}$ $\text{To find: the number of real roots.}$ $\text{Plan:}$ $1.\ \text{two cases for } |x^2-1|;\quad 2.\ \text{solve each};\quad 3.\ \text{check domain}$ $|x^2-1| = x^2-1 \ \text{ if } x^2 \ge 1$ $|x^2-1| = 1-x^2 \ \text{ if } x^2 < 1$ $\text{Case 1: } x \le -1 \ \text{or}\ x \ge 1$ $3x^2 - 4(x^2-1) + x - 1 = 0$ $-x^2 + x + 3 = 0 \Rightarrow x^2 - x - 3 = 0$ $x = \dfrac{1 \pm \sqrt{13}}{2} \approx 2.30,\ -1.30$ $\text{both satisfy } |x| \ge 1\ \checkmark$ $\text{Case 2: } -1 < x < 1$ $3x^2 - 4(1-x^2) + x - 1 = 0$ $7x^2 + x - 5 = 0$ $x = \dfrac{-1 \pm \sqrt{141}}{14} \approx 0.78,\ -0.92$ $\text{both lie in } (-1, 1)\ \checkmark$ $2 + 2 = 4\ \text{real roots}$ $\boxed{4}$