JEE Main 2025 — 7 April, Evening Shift — Question 38

$x|x-2|+3|x-3|+1=0$

Case I: $x<2$

$-x(x-2)-3(x-3)+1=0$

$-x^{2}+2 x-3 x+9+1=0$

$x^{2}+x-10=0$

$x=\frac{-1-\sqrt{41}}{2}$ or $\frac{-1+\sqrt{41}}{2}$ (rejected)

1 sol${ }^{\mathrm{n}}$.

Case II: $2 \leq x<3$ $x(x-2)-3(x-3)+1=0$

$x^{2}-5 x+8=0$

No solution

Case III: $x \geq 3$

$x(x-2)+3(x-3)+1=0$

$x^{2}+x-8=0$

$x=\frac{-1 \pm \sqrt{33}}{2}$ (no sol${ }^{\mathrm{n}}$ )

$\therefore x=\frac{-1-\sqrt{41}}{2}$ is the only solution