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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Quadratic Equation

The sum of all the real solutions of the equation _ (x+3) (6 x^ 2...

The sum of all the real solutions of the equation
log⁡(x+3)\log _{(x+3)}\left(6 x^{2}+28 x+30\right)=5-2 \log _{(6 x+10)}(x2+6x+9)\left(x^{2}+6 x+9\right) is equal to

Options

  1. A2
  2. B1
  3. C0
  4. D4

Correct answer

C. 0

Step-by-step solution

Note 6x2+28x+30=2(3x+5)(x+3)6x^2+28x+30 = 2(3x+5)(x+3),
x2+6x+9=(x+3)2x^2+6x+9 = (x+3)^2,
6x+10=2(3x+5)6x+10 = 2(3x+5).
Let a=log⁡(x+3)[2(3x+5)]a = \log_{(x+3)}[2(3x+5)]. Then LHS =a+1= a + 1 and log⁡2(3x+5)(x+3)2=2a\log_{2(3x+5)}(x+3)^2 = \frac{2}{a}.
Equation becomes a+1=5−4aa + 1 = 5 - \frac{4}{a}, i.e., a2−4a+4=0⇒a=2a^2 - 4a + 4 = 0 \Rightarrow a = 2.
So log⁡(x+3)[2(3x+5)]=2⇒(x+3)2=6x+10⇒x2=1⇒x=1\log_{(x+3)}[2(3x+5)] = 2 \Rightarrow (x+3)^2 = 6x+10 \Rightarrow x^2 = 1 \Rightarrow x = 1 or x=−1x = -1.
Both satisfy domain conditions (bases >0> 0, ≠1\neq 1; arguments >0> 0).
Sum =1+(−1)=0= 1 + (-1) = 0.
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