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JEE MainMathematicsQuadratic Equation

Let , be the roots of a quadratic equation x^2 - px + q = 0 , where p and q are positive real numbers. If S_n = ^n + ^n for n 1 , and it is given that S₂ = 14 and S₄ = 194 , then the value of S₅ is

Options

  1. A728
  2. B724
  3. C828
  4. D762

Correct answer

B. 724

Step-by-step solution

For the quadratic equation x^2 - px + q = 0 , we know that S₁ = + = p and = q . We can express S₂ and S₄ in terms of p and q : S₂ = ^2 + ^2 = ( + )^2 - 2 = p^2 - 2q S₄ = ^4 + ^4 = ( ^2 + ^2)^2 - 2 ^2 ^2 = S₂^2 - 2q^2 Given S₂ = 14 and S₄ = 194 , we substitute these into the second equation: 194 = 14^2 - 2q^2 194 = 196 - 2q^2 2q^2 = 2 q^2 = 1 Since q > 0 , we have q = 1 . Now substitute q = 1 and S₂ = 14 into the first equation: 14 = p^2 - 2(1) p^2 = 16 Since p > 0 , we have p = 4 . The quadratic equation is x^2 - 4

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