$x^2+b x+b=0$Here,a = 1b = bc = b$\therefore D = b^2 - 4 a c$$= b^2 - 4 \times 1 \times b$$= b^2 - 4 b$For the quadratic equation to have "two real and distinct roots".D >0$b^2 - 4 b > 0$$b(b-4) > 0$From the given options:-(A) 0b = 0$b(b-4) \ge 0$incorrect(B) 4b = 4$b(b-4) > 0$$4(4-4) > 0$$4 \times 0 > 0$0 > 0incorrect(C) 3b = 3b(b - 4) > 0$3(3-4) > 0$$3 \times (-1) > 0$-3 > 0incorrect.(D) -3b = -3$b(b-4) > 0$$-3(-3-4) > 0$$-3(-7) > 0$21 > 0