2022/031) The number of real roots of the equation $3x^2 - 4 |x^2-1| + x-1 = 0$ is

 there are 2 cases $x^2 >= 1$

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

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

or $x^2 -x  - 3 = 0$

this has two slutions one in $(1,\infty$ and another is $(-\infty, -1)$ 

for for the other case we have $3x^2 + 4(x^2-1) + x -1 = 0$ or $7x^2 + x - 5 = 0$ f(0) = -5, f(1) = 3 f(-1) = = 1 so there are 2 roots one between 0 and 1 and another between 0 and -1. 

So there are 4 solutions