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JEE Main202124 Feb 2021Morning ShiftMathematicsQuadratic EquationActual

Let p and q be two positive numbers such that p + q = 2 and p 4 + q 4 = 272 . Then p and q are roots of the equation:

Options

  1. Ax 2 - 2 x + 2 = 0
  2. Bx 2 - 2 x + 8 = 0
  3. Cx 2 - 2 x + 136 = 0
  4. Dx 2 - 2 x + 16 = 0

Correct answer

D. x 2 - 2 x + 16 = 0

Step-by-step solution

Consider p 2 + q 2 2 - 2 p 2 q 2 = 272 p + q 2 - 2 p q 2 - 2 p 2 q 2 = 272 16 - 16 p q + 2 p 2 q 2 = 272 p q 2 - 8 p q - 128 = 0 p q 2 - 8 p q - 128 = 0 p q = 8 ± 24 2 = 16 , - 8 ∴    p q = 16 ∴    Required equation :   x 2 - 2 x + 16 = 0

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