Question

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

Answer 1

Answer: 4

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