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

Solution

Case I

$x \le  – 1 \cup x \ge 1$

The equation becomes, $3x^2-4(x^2-1)+x-1=0$

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

$\Rightarrow x = \frac{{1 \mp \sqrt {13} }}{2}$

Both above values satisfy the condition $x \le  – 1 \cup x \ge 1$ and hence are accepted.

Case II

$-1 < x < 1$

The equation becomes, $3x^2-4(1-x^2)+x-1=0$

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

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

Both above values satisfy the condition $-1 < x < 1$ and hence are accepted.

So, 4 solutions in total.