# Question 3 [JEE adv, 2021]

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

## Concepts

Consider the sign of $$x^2 - 1$$ to get two quadratic equations.

## Solution

#### Step 1

$\begin{eqnarray} x^2 - 1 > 0&\implies& x\in (-\infty, -1)\cup(1,\infty)\\ x^2 - 1 < 0&\implies& x\in (-1, 1) \end{eqnarray}$

#### Step 2

##### Case 1: $$x^2 - 1>0$$

This gives the quadratic equation $\begin{eqnarray} 3x^2 -4(x^2 - 1) + x - 1 &=& 0\\ \implies x^2-x -3 &=& 0\\ \implies x &=& \frac{1\pm \sqrt{13}}{2} \end{eqnarray}$ Both the roots are in the allowed interval $$x\in (-\infty, -1)\cup(1,\infty)$$.

##### Case 2: $$x^2 - 1<0$$

This gives the quadratic equation $\begin{eqnarray} 3x^2 +4(x^2 - 1) + x - 1 &=& 0\\ \implies 7x^2+x -5 &=& 0\\ \implies x &=& \frac{-1\pm \sqrt{141}}{14} \end{eqnarray}$ Both the roots are in the allowed interval $$x\in (-1,1)$$.

#### Step 3

There are two roots each of the two quadratic equations. Hence, the total number of roots is 4.