For x ∈ R , the number of real roots of the equation 3 x 2 - 4 x 2 - 1 + x - 1 = 0 is

$\left|x^2-1\right|=\left\{\begin{array}{l}{x}^2-1, x^2\ge 1\\ -\left(x^2-1\right) x^2<1\end{array}\right.$

Case 1

$x^2-1<0\Rightarrow x\in (-1,1)$

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

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

$x=\frac{-1\pm \sqrt{1+140}}{2\times 7}=\frac{-1+\sqrt{141}}{14},\frac{-1-\sqrt{141}}{14}$

We know, $\sqrt{141}\approx 11.5$

So, both roots lie in the interval $\left(-1,1\right)$

Case 2

$x^2-1\ge 0\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )$

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

$-x^2+4+x-1=0$

$x^2-x-3=0$

$x=\frac{1\pm \sqrt{1+12}}{2}=\frac{1+\sqrt{13}}{2},\frac{1-\sqrt{13}}{2}$

We know, $3<\sqrt{13}<4$

So, both roots lie in the interval $(-\infty ,-1]\cup [1,\infty )$

Hence, $4$ real roots.