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
- A-81
- B100
- C144
- 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