Question

For $x \in R$, the number of real roots of the equation $3 x^{2}-4\left|x^{2}-1\right|+x-1=0$ is ______.

Answer 1

$3 x^{2}-4\left|x^{2}-1\right|+x-1=0$
Case-I: $|x| \geq 1 $
$3 x^{2}-4 x^{2}+4+x-1=0$
$-x^{2}+x+3=0$
$x^{2}-x-3=0 $
$x=\frac{1 \pm \sqrt{1+12}}{2}=\frac{1 \pm \sqrt{13}}{2}=\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2} $
$|x| < 1 $
$3 x^{2}+4 x^{2}-4+x-1=0 $
$7 x^{2}+x-5=0$
$x=\frac{-1 \pm \sqrt{1+20 \times 7}}{14}=\frac{-1 \pm \sqrt{141}}{14}$
So, number of real roots $=4$