1. **State the problem:** We need to prove that if the inequality $x^2 > k(x + 1)$ holds for all real $x$, then the constant $k$ must satisfy $-4 < k < 0$. 2. **Rewrite the inequality:** Move all terms to one side: $$x^2 - k(x + 1) > 0$$ which simplifies to $$x^2 - kx - k > 0$$ for all real $x$. 3. **Analyze the quadratic expression:** Consider the quadratic function $$f(x) = x^2 - kx - k$$ We want $f(x) > 0$ for every real $x$. 4. **Condition for positivity of quadratic:** A quadratic $ax^2 + bx + c$ with $a > 0$ is positive for all real $x$ if and only if its discriminant is negative: $$\Delta = b^2 - 4ac < 0$$ Here, $a = 1$, $b = -k$, and $c = -k$. 5. **Calculate the discriminant:** $$\Delta = (-k)^2 - 4(1)(-k) = k^2 + 4k$$ 6. **Set the discriminant condition:** $$k^2 + 4k < 0$$ 7. **Solve the inequality:** Factor: $$k(k + 4) < 0$$ This product is negative when $k$ is between the roots: $$-4 < k < 0$$ 8. **Conclusion:** For the inequality $x^2 > k(x + 1)$ to hold for all real $x$, the parameter $k$ must satisfy $$\boxed{-4 < k < 0}$$