Suppose $$k$$ is any integer such that the equation $$2x^2 + kx + 5 = 0$$ has no real roots, and the equation $$x^2 + (k - 5)x + 1 = 0$$ has two distinct real roots for $$x$$. Then, the number of possible values of $$k$$ is:

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#### Step 1: Analyze the First Equation The first equation is: $$ 2x^2 + kx + 5 = 0 $$ For this quadratic equation to have no real roots, its discriminant ($$D_1$$) must be negative: $$ D_1 = k^2 - 4(2)(5) $$ Simplify: $$ D_1 = k^2 - 40 $$ For no real roots: $$ k^2 - 40 4 $$ Solve the inequality: $$ |k - 5| > 2 $$ This gives two cases: 1. $$k - 5 > 2 \implies k > 7$$, 2. $$k - 5 < -2 \implies k < 3$$. Thus: $$ k \in (-\infty, 3) \cup (7, \infty) $$ --- #### Step 3: Combine Conditions From Step 1: $$k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$$. From Step 2: $$k \in (-\infty, 3) \cup (7, \infty)$$. The intersection of these two conditions is: $$ k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2\} $$ --- #### Step 4: Count the Values of $$k$$ The possible values of $$k$$ are: $$-6, -5, -4, -3, -2, -1, 0, 1, 2.$$ Thus, there are **9 possible values**. --- ### Final Answer: The number of possible values of $$k$$ is **9**, and the correct answer is **C)**.

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