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

$3x^2 - 4|x^2 - 1| + x - 1 = 0$

$x \in \mathbb{R}$

$\text{To find: number of real roots}$

|x² - 1| = x²-1, x² ≥ 1 1-x², x² < 1

$\text{Case 1: } x^2 \ge 1 \Rightarrow |x^2-1| = x^2-1$

$\text{Case 2: } x^2 < 1 \Rightarrow |x^2-1| = 1-x^2$

$\text{Case 1 } (x^2 \ge 1):$

$3x^2 - 4(x^2-1) + x - 1 = 0$

$-x^2 + x + 3 = 0$

$x^2 - x - 3 = 0$

$x = \dfrac{1 \pm \sqrt{13}}{2}$

$x \approx 2.303,\ -1.303\ (\text{both } |x| \ge 1)$

$\text{Case 2 } (x^2 < 1):$

$3x^2 - 4(1-x^2) + x - 1 = 0$

$7x^2 + x - 5 = 0$

$x = \dfrac{-1 \pm \sqrt{141}}{14}$

$x \approx 0.777,\ -0.920\ (\text{both } |x| < 1)$

$\text{Case 1: } 2 \text{ roots};\quad \text{Case 2: } 2 \text{ roots}$

$\text{Total real roots} = 2 + 2 = 4$

$\text{The graph meets the } x\text{-axis at } 4 \text{ points}$

$\boxed{4}$