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JEE Main201911 Jan 2019Morning ShiftMathematicsQuadratic EquationActual

If one real root of the quadratic equation 81 x²+k x+256=0 is cube of the other root, then a value of k is :

Options

  1. A-81
  2. B100
  3. C144
  4. D-300

Correct answer

D. -300

Step-by-step solution

Let and be the roots of the equation, 81 x²+k x+256=0 Given ( )^ 1 3 = = ³ Product of the roots = 256 81 ( )( )= 256 81 ⁴= ( 4 3 )⁴ = 4 3 = 64 27 Sum of the roots =- k 81 + =- k 81 4 3 + 64 27 =- k 81 k=-300

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