(B) 4
For two distinct real roots, discriminant $> 0$: $b^2 - 4(1)(b) > 0 \Rightarrow b^2 - 4b > 0 \Rightarrow b(b-4) > 0$, so $b < 0$ or $b > 4$. Only $b = 4$ does not satisfy… wait — checking $b = 4$: $16 - 16 = 0$ (equal roots). Checking $b = -3$: $9 - 4(-3) = 9 + 12 = 21 > 0$ ✓
Answer: (D) $-3$
Source: Chapter 4, Section 4.4 — Nature of Roots
For $x^2 + bx + b = 0$, here $a=1$, $c=b$, so discriminant $= b^2 - 4b$. For two distinct real roots, need $b^2 - 4b > 0$, i.e., $b(b-4) > 0$, which holds when $b < 0$ or $b > 4$. Among the options: $b=0$ gives $D=0$ (equal roots), $b=4$ gives $D=0$, $b=3$ gives $D=9-12=-3<0$ (no real roots), and $b=-3$ gives $D=9+12=21>0$ ✓. So the answer is D.