For which values of $a$ does the following equation have no real solution for $x:$

$$
\left(a^{2}+2 a\right) x^{2}+(3 a) x+1=0
$$
It is important to consider $a$ as a constant that does not change throughout the problem. We are reduced to a quadratic equation in the variable $x$. Whenever we have a quadratic equation in one variable, we solve it.

$$
\begin{gathered}
\Delta=(3 a)^{2}-4\left(a^{2}+2 a\right) \\
\Delta=5 a^{2}-10 a
\end{gathered}
$$

We want the equation in $x$ to have no solutions, so $\Delta<0$

$$
\begin{gathered}
5 a^{2}-10 a<0 \\
a(a-2)<0
\end{gathered}
$$

Therefore

$$
0<a<2
$$

Proposition 7 (Sum and Product of Roots). Consider an equation of the form $x^{2}-$ $S x+P=0$ where $S$ and $P$ are two real numbers. Suppose there are two roots (a root is a value of $x$ that makes the polynomial zero). Then $S$ is the sum, and $P$ is the product of these two roots.

It is indeed sufficient to calculate $x_{1}+x_{2}$ and $x_{1} x_{2}$ where $x_{1}=\frac{S+\sqrt{S^{2}-4 P}}{2}$ and $x_{2}=\frac{S-\sqrt{S^{2}-4 P}}{2}$.