3) The real number $a$ is such that the equation

$$
x^{2}+2 a x+1=0
$$

has two coincident real solutions. How many possible values of $a$ are there?
(A) None,
(B) one,
(C) two,
(D) three,
(E) four.
3. The answer is $(\mathbf{C})$.

A quadratic equation admits a single solution only in the case where the discriminant is equal to zero.

In our case, $\Delta=4 a^{2}-4$ must be zero. Therefore, we have $a^{2}-1=0$, which gives $a=1$ or $a=-1$. Thus, there are two values of $a$ for which the equation admits a single solution.