Multiple choice

Let $b$ be a non- zero real number. Suppose the quadratic equation $2x^2 + bx + \dfrac{1}{b} = 0$ has two distinct real roots. Then

  1. $b + \dfrac{1}{b} > \dfrac{5}{2}$
  2. $b + \dfrac{1}{b} < \dfrac{5}{2}$
  3. $b^2 - 3b > -2$
  4. $b^2 + \dfrac{1}{b^2} < 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the quadratic equation 2x^2 + bx + 1/b = 0 to have two distinct real roots, the discriminant D = b^2 - 4(2)(1/b) must be greater than 0. This simplifies to b^2 - 8/b > 0, which means (b^3 - 8)/b > 0. Testing values or analyzing the inequality shows that b^2 - 3b > -2 is a valid consequence of the root conditions.

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