Algebra – Quadratic Equations – JEE Main 24 Jan 2026 Shift 1

Solution:

We analyze the equation in different intervals defined by the critical points of the modulus functions: $x=-3$ and $x=1$.

**Case 1: $x < -3$** Both $|x+3|$ and $|x-1|$ open with negative signs. $$x(-(x+3)) - (x-1) - 2 = 0$$ $$-x^2 - 3x - x + 1 - 2 = 0$$ $$-x^2 - 4x - 1 = 0 \Rightarrow x^2 + 4x + 1 = 0$$ Roots: $x = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$. $x_1 = -2 + 1.732 = -0.268$ (Rejected, as $x < -3$) $x_2 = -2 - 1.732 = -3.732$ (Accepted) $\Rightarrow$ 1 Solution.
**Case 2: $-3 \le x < 1$** $|x+3|$ opens positive, $|x-1|$ opens negative. $$x(x+3) - (x-1) - 2 = 0$$ $$x^2 + 3x - x + 1 - 2 = 0$$ $$x^2 + 2x - 1 = 0$$ Roots: $x = \frac{-2 \pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}$. $x_3 = -1 + 1.414 = 0.414$ (Accepted, lies in $[-3, 1)$) $x_4 = -1 - 1.414 = -2.414$ (Accepted, lies in $[-3, 1)$) $\Rightarrow$ 2 Solutions.
**Case 3: $x \ge 1$**
Both moduli open positive.
$$x(x+3) + (x-1) – 2 = 0$$
$$x^2 + 3x + x – 3 = 0$$
$$x^2 + 4x – 3 = 0$$
Roots: $x = \frac{-4 \pm \sqrt{16+12}}{2} = -2 \pm \sqrt{7}$.
$x_5 = -2 + 2.64 = 0.64$ (Rejected, as $x \ge 1$)
$x_6 = -2 – 2.64 = -4.64$ (Rejected, as $x \ge 1$)
$\Rightarrow$ 0 Solutions.

Total number of solutions = $1 + 2 + 0 = 3$.

Ans. (1)