Basics & LogarithmsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Absolute Value Equation Solution Set: Sum = 18 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If the set of all solutions of ∣x2+x−9∣=∣x∣+∣x2−9∣|x^2+x-9|=|x|+|x^2-9| is [α,β]∪[γ,∞)[\alpha,\beta]\cup[\gamma,\infty), then (α2+β2+γ2)(\alpha^2+\beta^2+\gamma^2) is equal to
A99
B1818correct
C3636
D7272
Solution
Step 1: The identity ∣a∣+∣b∣=∣a+b∣|a|+|b|=|a+b| holds iff ab≥0ab\ge0. Here a=xa=x and b=x2−9b=x^2-9, so we need
x(x2−9)≥0.x(x^2-9)\ge0.
 Step 2: Solve x(x−3)(x+3)≥0x(x-3)(x+3)\ge0: the solution is
x∈[−3,0]∪[3,∞).x\in[-3,0]\cup[3,\infty).
 Step 3: Matching to [α,β]∪[γ,∞)[\alpha,\beta]\cup[\gamma,\infty): α=−3, β=0, γ=3\alpha=-3,\ \beta=0,\ \gamma=3. Then
α2+β2+γ2=9+0+9=18.\alpha^2+\beta^2+\gamma^2=9+0+9=18.
 Correct answer: (2)
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