Algebra – Quadratic Equations – JEE Main 21 Jan 2026 Shift 1

Solution:

Let $|x-1| = t$. Since $(x-1)^2 = |x-1|^2$, the equation can be written as:
$$t^2 – 5t + 6 = 0$$

Factorizing the quadratic equation:
$$(t-2)(t-3) = 0$$
So, $t = 2$ or $t = 3$.

Case 1: $|x-1| = 2$
$$x – 1 = \pm 2$$
$$x = 1 + 2 = 3 \quad \text{or} \quad x = 1 – 2 = -1$$

Case 2: $|x-1| = 3$
$$x – 1 = \pm 3$$
$$x = 1 + 3 = 4 \quad \text{or} \quad x = 1 – 3 = -2$$

The set of roots is $\{3, -1, 4, -2\}$.
Sum of all roots:
$$\text{Sum} = 3 + (-1) + 4 + (-2)$$
$$\text{Sum} = 7 – 3 = 4$$

Ans. (1)