JEE Main202623 January 2026Morning ShiftMathematicsQuadratic EquationActual
If and ( < ) are the roots of the equation (-2+ 3 )(| x -3|)+(x-6 x )+(9-2 3 )=0, x 0 , then + is equal to :
Options
- A8
- B11
- C9
- D10
Correct answer
D. 10
Step-by-step solution
Let t = x 0 . The equation becomes (-2+ 3 )|t-3| + (t^2 - 6t) + (9 - 2 3 ) = 0 . Since t^2 - 6t + 9 = (t-3)^2 , we get (t-3)^2 + ( 3 -2)|t-3| - 2 3 = 0 . Let u = |t-3| 0 : u^2 + ( 3 -2)u - 2 3 = 0 . Discriminant = ( 3 -2)^2 + 8 3 = 7 + 4 3 = (2+ 3 )^2 . u = (2- 3 ) (2+ 3 ) 2 , giving u = 2 or u = - 3 (rejected). |t-3| = 2 t = 5 or t = 1 , so x = 25 or x = 1 . = 1, = 25 . / + = 25 + 25 = 5 + 5 = 10 .