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(C) Given the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$,it has two distinct real roots,so the discriminant $D > 0$.$D = b^2 - 4(2)(\frac{1}{b})

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$b^2 - 3b > -2$

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(C) Given the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$,it has two distinct real roots,so the discriminant $D > 0$.$D = b^2 - 4(2)(\frac{1}{b}) > 0$$b^2 - \frac{8}{b} > 0 \Rightarrow \frac{b^3 - 8}{b} > 0$Using the factorization $b^3 - 8 = (b - 2)(b^2 + 2b + 4)$,and noting that $b^2 + 2b + 4 = (b + 1)^2 + 3 > 0$ for all real $b$,the inequality simplifies to:$\frac{b - 2}{b} > 0$This holds when $b \in (-\infty, 0) \cup (2, \infty)$.Now,check option $(c)$:$b^2 - 3b > -2 \Rightarrow b^2 - 3b + 2 > 0$$(b - 2)(b - 1) > 0$This holds when $b \in (-\infty, 1) \cup (2, \infty)$.Since the set $(-\infty, 0) \cup (2, \infty)$ is a subset of $(-\infty, 1) \cup (2, \infty)$,the condition $b^2 - 3b > -2$ is satisfied for all $b$ that satisfy the original quadratic equation condition.

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

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