3. Given that $a$, $b$, and $c$ are positive numbers, the quadratic equation $a x^{2}+b x+c=0$ has two equal real roots. Then the root situation of the equation $(a+1) x^{2}+(b+2) x+(c+1)=0$ is $(\quad)$.
(A) No real roots
(B) Two equal real roots
(C) Two unequal real roots
(D) The root situation is uncertain
3. D.

Let the equal root of $a x^{2}+b x+c=0$ be $r$. Then
$$
a x^{2}+b x+c=a(x-r)^{2} \text {. }
$$

When $r=-1$,
$$
\begin{array}{l}
(a+1) x^{2}+(b+2) x+(c+1) \\
=\left(a x^{2}+b x+c\right)+(x+1)^{2}=(a+1)(x+1)^{2},
\end{array}
$$

it can be known that $x=-1$ is also the equal root of $(a+1) x^{2}+(b+2) x+(c+1)=0$;
When $r \neq-1$, for any real number $x$, we have
$$
\begin{array}{l}
(a+1) x^{2}+(b+2) x+(c+1) \\
=a(x-r)^{2}+(x+1)^{2}>0,
\end{array}
$$

which means the equation has no real roots.