Number of Real Roots for Equation with Absolute Value

To find the total number of real roots for the given equation $3x^2 - 4|x^2 - 1| + x - 1 = 0$ we analyze the absolute value expression $|x^2 - 1|$ by considering two cases based on the sign of $x^2 - 1$.

---

Case 1: $x^2 - 1 \ge 0$

This condition corresponds to $x \in (-\infty, -1] \cup [1, \infty)$.
Here, $|x^2 - 1| = x^2 - 1$. Substituting this into the given equation gives: $3x^2 - 4(x^2 - 1) + x - 1 = 0$ $-x^2 + x + 3 = 0 \implies x^2 - x - 3 = 0$

Using the quadratic formula, the roots are: $x = \frac{1 \pm \sqrt{1 - 4(1)(-3)}}{2} = \frac{1 \pm \sqrt{13}}{2}$

Now, we verify if these roots lie in the domain $(-\infty, -1] \cup [1, \infty)$:

1. For $x_1 = \frac{1 + \sqrt{13}}{2}$:
   Since $\sqrt{13} > 3$, $x_1 > \frac{1 + 3}{2} = 2 \ge 1$. Thus, $x_1 \in [1, \infty)$.

2. For $x_2 = \frac{1 - \sqrt{13}}{2}$:
   Since $\sqrt{13} \ge 3$, $x_2 \le \frac{1 - 3}{2} = -1$. Thus, $x_2 \in (-\infty, -1]$.

Both roots are valid solutions. So, Case 1 yields 2 real roots.

---

Case 2: $x^2 - 1 < 0$

This condition corresponds to $x \in (-1, 1)$.
Here, $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$. Substituting this into the given equation gives: $3x^2 - 4(1 - x^2) + x - 1 = 0$ $3x^2 - 4 + 4x^2 + x - 1 = 0$ $7x^2 + x - 5 = 0$

Using the quadratic formula, the roots are: $x = \frac{-1 \pm \sqrt{1 - 4(7)(-5)}}{14} = \frac{-1 \pm \sqrt{141}}{14}$

Now, we verify if these roots lie in the domain $(-1, 1)$:

1. For $x_3 = \frac{-1 + \sqrt{141}}{14}$:
   Since $11 < \sqrt{141} < 12$, we have: $0 < \frac{-1 + 11}{14} < x_3 < \frac{-1 + 12}{14} < 1$ Thus, $x_3 \in (-1, 1)$.

2. For $x_4 = \frac{-1 - \sqrt{141}}{14}$:
   Since $11 < \sqrt{141} < 13$, we have: $-1 = \frac{-1 - 13}{14} < x_4 < \frac{-1 - 11}{14} < 0$ Thus, $x_4 \in (-1, 1)$.

Both roots are valid solutions. So, Case 2 yields 2 real roots.

---

Conclusion:

Combining both cases, the equation has $2 + 2 = 4$ distinct real roots.

The total number of real roots is 4.