For $$x \in R$$, the number of real roots of the equation
$$3x^2 - 4 \mid x^2 - 1 \mid + x - 1 = 0$$ is .........

The given equation contains an absolute‐value term, so we first remove the modulus by considering the sign of the expression inside it.

The modulus appears as $$\lvert x^{2}-1 \rvert$$. Split the real line into the two natural regions:

Case 1: $$\lvert x \rvert \ge 1 \;(\;x^{2}\ge 1\;)$$
Then $$\lvert x^{2}-1\rvert = x^{2}-1$$ because the quantity inside the modulus is non-negative.

Substituting $$\lvert x^{2}-1\rvert = x^{2}-1$$ in the original equation

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

and simplifying:

$$3x^{2}-4x^{2}+4+x-1=0$$
$$-x^{2}+x+3=0$$
Multiply by $$-1$$ to obtain a standard quadratic:

$$x^{2}-x-3=0 \;-(1)$$

For $$ax^{2}+bx+c=0$$ the discriminant is $$\Delta=b^{2}-4ac$$.
Here $$a=1,\,b=-1,\,c=-3$$, so $$\Delta=(-1)^{2}-4(1)(-3)=1+12=13$$.

Since $$\Delta\gt 0$$ the quadratic has two real roots:

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

Numerically, $$\frac{1+\sqrt{13}}{2}\approx 2.303,\; \frac{1-\sqrt{13}}{2}\approx -1.303$$. Both satisfy $$\lvert x\rvert \ge 1$$, so both are valid roots in Case 1.

Case 2: $$\lvert x \rvert \lt 1 \;(\;x^{2}\lt 1\;)$$
Now $$\lvert x^{2}-1\rvert = 1-x^{2}$$ because the quantity inside the modulus is negative.

Substituting $$\lvert x^{2}-1\rvert = 1-x^{2}$$ in the equation

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

Simplify term by term:

$$3x^{2}-4+4x^{2}+x-1=0$$
$$7x^{2}+x-5=0 \;-(2)$$

Again compute the discriminant: $$a=7,\,b=1,\,c=-5$$

$$\Delta = 1^{2}-4(7)(-5)=1+140=141$$

With $$\Delta\gt 0$$ we have two real roots:

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

Approximate values: $$\frac{-1+\sqrt{141}}{14}\approx 0.777,\; \frac{-1-\sqrt{141}}{14}\approx -0.920$$.
Both lie strictly between $$-1$$ and $$1$$, so both satisfy $$\lvert x\rvert\lt 1$$ and are valid roots in Case 2.

Checking the boundary points $$x=\pm1$$ (where the definition of the modulus changes):
For $$x=1$$: $$3(1)^{2}-4\cdot 0+1-1=3\neq 0$$.
For $$x=-1$$: $$3(1)-4\cdot0-1-1=1\neq 0$$.
Hence neither boundary point is a solution.

Total number of distinct real roots = 2 (from Case 1) + 2 (from Case 2) = 4.

Final answer: 4 real roots.