Verified Solution • 2025–26 Session moderate

QJEEM19Ma0518: JEE Main 2019 Mathematics - Class 11 JEEM19Ma0518 Solution: If one real root of the quadratic equation 81x 2 + kx + 256 = 0 …

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

Multiple Choice Options

  • A −- 81
  • B −- 300
  • C 100
  • D 144

Step-by-Step Verified Solution

81x2 + kx + 256 = 0 ; x = α\alpha , α\alpha 3

⇒\Rightarrow    α\alpha 4 = 25681{{256} \over {81}} ⇒\Rightarrow    α\alpha = ±\pm 43{{4} \over {3}}

Now    −- k81{k \over {81}} = α\alpha + α\alpha 3 = ±\pm 10027{{100} \over {27}}

⇒\Rightarrow   k = ±\pm 300

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