Find all values of the parameter $a$ such that the equation

$$
a x^{2}-(a+3) x+2=0
$$

admits two real roots of opposite signs.

Definition. Let $n \geq 1$ be an integer. Let $z_{1}, \ldots, z_{n} \in \mathbb{K}$. For $1 \leq k \leq n$, the $k$-th elementary symmetric function is defined as

$$
\sigma_{k}\left(z_{1}, \ldots, z_{n}\right)=\sum_{1 \leq i_{1}<\cdots<i_{k} \leq n} z_{i_{1}} \cdots z_{i_{k}}
$$

When the elements $z_{1}, \ldots, z_{n}$ are understood, for simplicity, we sometimes write $\sigma_{k}$ instead of $\sigma_{k}\left(z_{1}, \ldots, z_{n}\right)$.

Thus, for example, for $n=3$, we have $\sigma_{1}=z_{1}+z_{2}+z_{3}, \sigma_{2}=z_{1} z_{2}+z_{1} z_{3}+z_{2} z_{3}$ and $\sigma_{3}=z_{1} z_{2} z_{3}$.

Proposition (Vieta's formulas in the general case). Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+$ $\cdots+a_{1} x+a_{0} \in \mathbb{K}[X]$ with $a_{n} \neq 0$. Suppose that $P$ has $n$ roots $z_{1}, \ldots, z_{n} \in \mathbb{K}$ counted with multiplicity (i.e., if $z$ is a root of order $k, z$ appears $k$ times in the list $z_{1}, \ldots, z_{n}$). Then, for any integer $1 \leq k \leq n$,

$$
\sigma_{k}\left(z_{1}, \ldots, z_{k}\right)=(-1)^{k} \frac{a_{n-k}}{a_{n}}
$$

In other words,

$$
\sum_{i=1}^{n} z_{i}=-\frac{a_{n-1}}{a_{n}}, \quad \sum_{1 \leq i<j \leq n} z_{i} z_{j}=-\frac{a_{n-2}}{a_{n}}, \ldots, \prod_{i=1}^{n} z_{i}=(-1)^{n} \frac{a_{0}}{a_{n}}
$$

Proof. According to the Corollary, we have $P(x)=a_{n}\left(x-z_{1}\right) \cdots\left(x-z_{n}\right)$. In the expansion of $a_{n}\left(x-z_{1}\right) \cdots\left(x-z_{n}\right)$, it is easy to see that the coefficient of $x^{n-k}$ is $(-1)^{k} a_{n} \sigma_{k}\left(z_{1}, \ldots, z_{k}\right)$. This coefficient is also equal to $a_{n-k}$. The result follows.

Example. Find all real numbers $x, y, z$ such that

$$
x+y+z=17, \quad x y+y z+x z=94, \quad x y z=168 .
$$

According to Vieta's formulas, $x, y, z$ are roots of $P(x)=x^{3}-17 x^{2}+$ $94 x-168=0$. Let's look for "obvious" roots of $P$. It is natural to first look for integer roots, which must be divisors of 168 (to test if, for example, 2 is a root, we perform the Euclidean division of $P$ by $x-2$ and check if the remainder is zero). We notice that $x=4$ is a root, and without difficulty, we find that $x=6$ and $x=7$ are roots of the polynomial $P(x) /(x-4)$. Thus, the solutions $(x, y, z)$ are the six possible permutations of $(4,6,7)$.

Remark. The functions $\sigma_{1}\left(z_{1}, \ldots, z_{n}\right), \ldots, \sigma_{n}\left(z_{1}, \ldots, z_{n}\right)$ are called the elementary symmetric functions of the $z_{i}$. Symmetric, because a permutation of the $z_{i}$ leaves the $\sigma_{k}$ invariant. Elementary, because it can be shown that any symmetric expression in $n$ variables can be expressed polynomially using these elementary symmetric functions. More precisely, if $P\left(z_{1}, \ldots, z_{n}\right)$ is a polynomial in $n$ variables (the reader is left to imagine what this is) such that for any permutation $\sigma$ of $\{1, \ldots, n\}$ we have $P\left(z_{1}, \ldots, z_{n}\right)=P\left(z_{\sigma(1)}, \ldots, z_{\sigma(n)}\right)$, then there exists a polynomial in $n$ variables $R$ such that $P\left(z_{1}, \ldots, z_{n}\right)=R\left(\sigma_{1}\left(z_{1}, \ldots, z_{n}\right), \ldots, \sigma_{n}\left(z_{1}, \ldots, z_{n}\right)\right)$.

Example. Denoting $\sigma_{1}=x_{1}+x_{2}+x_{3}, \sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}$ and $\sigma_{3}=x_{1} x_{2} x_{3}$, we have:

$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=\sigma_{1}^{3}-3 \sigma_{1} \sigma_{2}+3 \sigma_{3}
$$

In short, when dealing with symmetric quantities, it can sometimes be useful to involve the associated elementary symmetric functions.
Suppose that $a x^{2}-(a+3) x+2=0$ has two roots of opposite sign, denoted by $z_{1}, z_{2}$. Then, according to Vieta's formulas, $z_{1} z_{2} = 2 / a$. Now, $z_{1}$ and $z_{2}$ are of opposite sign if and only if $z_{1} z_{2} < 0$. We deduce that $a < 0$. Conversely, if $a < 0$, then the discriminant of the equation is $a^{2}-2 a+9$. To show that it is positive, we use the canonical form by writing $a^{2}-2 a+9=(a-1)^{2}+8 \geq 0$. Thus, when $a < 0$, there are two real solutions denoted by $z_{1}, z_{2}$. According to Vieta's formulas, $z_{1} z_{2}=2 / a < 0$, so $z_{1}$ and $z_{2}$ are of opposite sign.

Note that in the proof of the converse, it was first necessary to show that the polynomial has two real roots before using Vieta's formulas.