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KVPY2019MathematicsQuadratic Equation

Let b be an non-zero real number. Suppose the quadratic equation 2 x 2 + b x + 1 b = 0 has two distinct real roots. Then

Options

  1. Ab + 1 b > 5 2
  2. Bb + 1 b < 5 2
  3. Cb 2 - 3 b > - 2
  4. Db 2 + 1 b 2 < 4

Correct answer

C. b 2 - 3 b > - 2

Step-by-step solution

Given quadratic equation 2 x 2 + b x + 1 b = 0 , has two distinct real roots, so ⇒ D > 0 b 2 - 4 ( 2 ) 1 b > 0 ⇒    b 2 - 8 b > 0 ⇒ b 3 - 8 b > 0 ⇒    ( b - 2 ) b 2 + 2 b + 4 b > 0 ⇒ b ∈ ( - ∞ , 0 ) ∪ ( 2 , ∞ ) For option (c), b 2 - 3 b > - 2 ⇒    b 2 - 3 b + 2 > 0 ⇒ ( b - 2 ) ( b - 1 ) > 0 b ∈ ( - ∞ , 1 ) ∪ ( 2 , ∞ ) mean if b ∈ ( - ∞ , 0 ) ∪ ( 2

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