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JEE Main202623 January 2026Evening ShiftMathematicsQuadratic EquationActual

The sum of all the real solutions of the equation _ (x+3) (6 x²+28 x+30 )=5-2 _ (6 x+10) (x²+6 x+9 ) is equal to

Options

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

Correct answer

C. 0

Step-by-step solution

Note 6x^2+28x+30 = 2(3x+5)(x+3) , x^2+6x+9 = (x+3)^2 , 6x+10 = 2(3x+5) . Let a = _ (x+3) [2(3x+5)] . Then LHS = a + 1 and _ 2(3x+5) (x+3)^2 = 2 a . Equation becomes a + 1 = 5 - 4 a , i.e., a^2 - 4a + 4 = 0 a = 2 . So _ (x+3) [2(3x+5)] = 2 (x+3)^2 = 6x+10 x^2 = 1 x = 1 or x = -1 . Both satisfy domain conditions (bases > 0 , 1 ; arguments > 0 ). Sum = 1 + (-1) = 0 .

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