Find the number of real roots of the equation 3x² - 4|x² - 1| + x - 1 = 0 — JEE Mathematics

We analyze the expression inside the absolute value, $|x^2 - 1|$. This changes behavior at $x = \pm 1$.

Case 1: $|x| \ge 1 \implies x^2 - 1 \ge 0$
The equation becomes:

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

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

Using the quadratic formula:

$$x = \frac{1 \pm \sqrt{1 - 4(1)(-3)}}{2} = \frac{1 \pm \sqrt{13}}{2}$$

Since $\sqrt{13} \approx 3.6$, the roots are $\approx 2.3$ and $\approx -1.3$. Both satisfy $|x| \ge 1$. (2 real roots)

Case 2: $|x| < 1 \implies x^2 - 1 < 0$
The equation becomes:

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

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

Using the quadratic formula:

$$x = \frac{-1 \pm \sqrt{1 - 4(7)(-5)}}{14} = \frac{-1 \pm \sqrt{141}}{14}$$

Since $\sqrt{141} \approx 11.87$, the roots are $\approx \frac{10.87}{14} \approx 0.77$ and $\approx \frac{-12.87}{14} \approx -0.92$. Both roots satisfy $|x| < 1$. (2 real roots)

Total number of real roots = $2 + 2 = 4$.

Answer: 4