Find the range of $k$ such that the inequality $x^2 + (k+3)x + k+3 > 0$ holds for all real numbers $x$.

1. Problem Description

Find the range of $k$ such that the inequality $x^2 + (k+3)x + k+3 > 0$ holds for all real numbers $x$.

2. Solution Steps

For the quadratic inequality $ax^2 + bx + c > 0$ to be true for all real numbers $x$, two conditions must be met:

1. The coefficient of $x^2$ must be positive, i.e., $a > 0$.

2. The discriminant must be negative, i.e., $b^2 - 4ac < 0$.

In our case, $a = 1$, $b = k+3$, and $c = k+3$.

The coefficient of $x^2$ is $1$, which is positive. So the first condition is satisfied.

Now we need to satisfy the second condition: the discriminant must be negative.

The discriminant is

$D = b^2 - 4ac = (k+3)^2 - 4(1)(k+3) < 0$.

$(k+3)^2 - 4(k+3) < 0$

$(k+3)(k+3 - 4) < 0$

$(k+3)(k-1) < 0$.

Now we analyze the inequality $(k+3)(k-1) < 0$.

The roots are $k = -3$ and $k = 1$.

We consider three intervals: $k < -3$, $-3 < k < 1$, and $k > 1$.

If $k < -3$, then $k+3 < 0$ and $k-1 < 0$, so $(k+3)(k-1) > 0$.

If $-3 < k < 1$, then $k+3 > 0$ and $k-1 < 0$, so $(k+3)(k-1) < 0$.

If $k > 1$, then $k+3 > 0$ and $k-1 > 0$, so $(k+3)(k-1) > 0$.

Thus, the inequality $(k+3)(k-1) < 0$ is satisfied when $-3 < k < 1$.

3. Final Answer

The range of $k$ is $-3 < k < 1$.