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Basics & Logarithms: Find Modulus Sum Solutions

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

Question
Find the modulus of the sum of all solutions to (x2+3x+1)x2+2x−8=1(x^2+3x+1)^{x^2+2x-8} = 1 is
Solution
Answer: 7
An exponential expression BE=1B^E = 1 when: (i) E=0E = 0 and B≠0B \neq 0, (ii) B=1B = 1, or (iii) B=−1B = -1 and EE is even. Case (i): Exponent =0= 0:
x2+2x−8=0  ⟹  (x+4)(x−2)=0  ⟹  x=−4 or x=2x^2+2x-8=0 \implies (x+4)(x-2)=0 \implies x=-4 \text{ or } x=2
 Check bases: x=−4x=-4: B=16−12+1=5≠0B=16-12+1=5\neq0. Valid. x=2x=2: B=4+6+1=11≠0B=4+6+1=11\neq0. Valid. Case (ii): Base =1= 1:
x2+3x+1=1  ⟹  x2+3x=0  ⟹  x(x+3)=0  ⟹  x=0 or x=−3x^2+3x+1=1 \implies x^2+3x=0 \implies x(x+3)=0 \implies x=0 \text{ or } x=-3
 Both valid. Case (iii): Base =−1= -1 and exponent even:
x2+3x+1=−1  ⟹  x2+3x+2=0  ⟹  (x+1)(x+2)=0  ⟹  x=−1 or x=−2x^2+3x+1=-1 \implies x^2+3x+2=0 \implies (x+1)(x+2)=0 \implies x=-1 \text{ or } x=-2
 x=−1x=-1: exponent =1−2−8=−9=1-2-8=-9 (odd). (−1)−9=−1≠1(-1)^{-9}=-1\neq1. Rejected. x=−2x=-2: exponent =4−4−8=−8=4-4-8=-8 (even). (−1)−8=1(-1)^{-8}=1. Valid. All solutions: {−4, 2, 0, −3, −2}\{-4,\, 2,\, 0,\, -3,\, -2\}. Sum =−4+2+0−3−2=−7= -4+2+0-3-2 = -7. ∣−7∣=7|-7| = 7. Answer: 7
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