JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

The sum of all the roots of the equation $(x - 1)^{2} - 5∣ x - 1∣ + 6 = 0$ , is:

Solution

Let $t = ∣ x - 1∣$ (so $t \geq 0$ ). Then $t^{2} - 5 t + 6 = 0 \Rightarrow (t - 2) (t - 3) = 0$ , giving $t = 2$ or $t = 3$ .

$∣ x - 1∣ = 2 \Rightarrow x = 3$ or $- 1$ . $∣ x - 1∣ = 3 \Rightarrow x = 4$ or $- 2$ .

Sum of roots $= 3 + (- 1) + 4 + (- 2) = 4$ .

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.