JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Quadratic Equations and Inequalities

Rewrite as $3x^2+x-1=4|x^2-1|$. For $x\in [-1,1]$: $3x^2+x-1=-4x^2+4$, giving $7x^2+x-5=0$. Let $f(x)=7x^2+x-5$. We have $f(1)=3>0$, $f(-1)=1>0$, $f(0)=-5<0$. By the intermediate value theorem, there are two roots in $[-1,1]$. For $x\in (-\infty ,-1]\cup [1,\infty )$: $3x^2+x-1=4x^2-4$, giving $x^2-x-3=0$. Let $g(x)=x^2-x-3$. We have $g(-1)=-1<0$ and $g(1)=-3<0$. Since the discriminant is positive and the parabola opens upward, there are two roots outside $[-1,1]$. Total: 4 roots.