JEE Advanced 2021 Paper 1 · Q17 · Quadratic Equations

Question

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

SolveKar AI · Accepted Answer

Case I: $|x|\ge 1$, so $|x^2-1|=x^2-1$. The equation becomes $3x^2-4(x^2-1)+x-1=0\Rightarrow -x^2+x+3=0\Rightarrow x^2-x-3=0\Rightarrow x=\dfrac{1\pm\sqrt{13}}{2}$. Both values satisfy $|x|\ge 1$ since $\dfrac{1+\sqrt{13}}{2}\approx 2.30$ and $\dfrac{1-\sqrt{13}}{2}\approx -1.30$. So 2 roots. Case II: $|x|<1$, $|x^2-1|=1-x^2$. The equation becomes $3x^2+4(x^2-1)+x-1$... actually $3x^2-4(1-x^2)+x-1=0\Rightarrow 7x^2+x-5=0\Rightarrow x=\dfrac{-1\pm\sqrt{141}}{14}$. $\sqrt{141}\approx 11.87$, so $x\approx 0.776$ or $x\approx -0.919$; both have $|x|<1$. So 2 more roots. Total = 4.

JEE Advanced 2021 Paper 1 · Q17 · Quadratic Equations

Solution