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Basics & Logarithms: Number Values Satisfying

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

Question
Number of values of xx satisfying ∣x2+4x+3∣+2x+5=0|x^2 + 4x + 3| + 2x + 5 = 0 is
Solution
Answer: 2
Step 1: Note x2+4x+3=(x+1)(x+3)x^2+4x+3 = (x+1)(x+3). Critical points: x=−3x = -3 and x=−1x = -1. For the equation to hold, 2x+5≤02x+5 \leq 0, i.e., x≤−52x \leq -\dfrac{5}{2}. Case 1: x≤−3x \leq -3 (so x2+4x+3≥0x^2+4x+3 \geq 0):
(x2+4x+3)+(2x+5)=0  ⟹  x2+6x+8=0  ⟹  (x+4)(x+2)=0(x^2+4x+3) + (2x+5) = 0 \implies x^2+6x+8=0 \implies (x+4)(x+2)=0
 x=−4x = -4 (satisfies x≤−3x \leq -3, valid) or x=−2x = -2 (rejected). Case 2: −3<x≤−52-3 < x \leq -\dfrac{5}{2} (so x2+4x+3<0x^2+4x+3 < 0):
−(x2+4x+3)+(2x+5)=0  ⟹  x2+2x−2=0  ⟹  x=−1±3-(x^2+4x+3) + (2x+5) = 0 \implies x^2+2x-2=0 \implies x = -1 \pm \sqrt{3}
 x=−1−3≈−2.73x = -1-\sqrt{3} \approx -2.73, which satisfies −3<x≤−2.5-3 < x \leq -2.5. Valid. x=−1+3>0x = -1+\sqrt{3} > 0, rejected. Step 2: Two solutions: x=−4x = -4 and x=−1−3x = -1-\sqrt{3}. Answer: 2
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