Suppose $k$ is any integer such that the equation $2 x^{2}+k x+5=0$ 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

We need to find integer values of $k$ such that:

1. $2x^2 + kx + 5 = 0$ has no real roots
2. $x^2 + (k-5)x + 1 = 0$ has two distinct real roots

For any quadratic equation $ax^2 + bx + c = 0$, the discriminant $D = b^2 - 4ac$ tells us about the nature of roots:

If $D < 0$: No real roots

If $D = 0$: One repeated real root

If $D > 0$: Two distinct real roots

---

For $2x^2 + kx + 5 = 0$:

Here, $a = 2$, $b = k$, $c = 5$

Discriminant: $D_1 = k^2 - 4(2)(5) = k^2 - 40$

For no real roots: $D_1 < 0$

$k^2 - 40 < 0$

$k^2 < 40$

Since $\sqrt{40} = 2\sqrt{10} \approx 6.32$, we get:

$-6.32 < k < 6.32$

Since $k$ must be an integer: $k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$

---

For $x^2 + (k-5)x + 1 = 0$:

Here, $a = 1$, $b = (k-5)$, $c = 1$

Discriminant: $D_2 = (k-5)^2 - 4(1)(1) = (k-5)^2 - 4$

For two distinct real roots: $D_2 > 0$

$(k-5)^2 - 4 > 0$

$(k-5)^2 > 4$

When $(k-5)^2 > 4$, we have:

$|k-5| > 2$

This absolute value inequality gives us two cases:

Case 1: $k-5 > 2 \rightarrow k > 7$

Case 2: $k-5 < -2 \rightarrow k < 3$

So from condition 2: $k < 3$ or $k > 7$

---

We need values of $k$ that satisfy both conditions:

Condition 1: $-6 \leq k \leq 6$ (integers)

Condition 2: $k < 3$ or $k > 7$

The range $k > 7$ doesn't overlap with $-6 \leq k \leq 6$

So we only consider $k < 3$ combined with $-6 \leq k \leq 6$

This gives us: $-6 \leq k < 3$ where $k$ is an integer.

Therefore: $k \in \{-6, -5, -4, -3, -2, -1, 0, 1, 2\}$

---

The number of possible values of $k$ is 9.

Note that $k = 3$ would make the second equation have exactly one repeated root (since $D_2 = 0$), which violates our "two distinct roots" condition.