4. Determine all pairs $a, b$ of real numbers for which each of the quadratic equations

$$
a x^{2}+2 b x+1=0, \quad b x^{2}+2 a x+1=0
$$

has two distinct real roots, and exactly one of them is common to both equations.
SOLUTION. From the problem statement, it follows that $a \neq 0, b \neq 0$ (the equations would not be quadratic) and $a \neq b$ (the equations would be identical, and if they had two real roots, they would both be common).

Let $x_{0}$ be the common root of both equations, so that

$$
a x_{0}^{2}+2 b x_{0}+1=0, \quad b x_{0}^{2}+2 a x_{0}+1=0 .
$$

By subtracting the two equations, we get $(a-b)\left(x_{0}^{2}-2 x_{0}\right)=x_{0}(a-b)\left(x_{0}-2\right)=0$. Since $a \neq b$ and 0 is clearly not a root of the given equations, the common root must be $x_{0}=2$. Substituting into the given equations, we then get the single condition $4 a+4 b+1=0$, or

$$
b=-a-\frac{1}{4}
$$

The discriminant of the second of the given equations is then $4 a^{2}-4 b=4 a^{2}+4 a+1=(2 a+1)^{2}$, so the equation has two distinct real roots for any $a \neq-\frac{1}{2}$. Similarly, the discriminant of the first of the given equations is $4 b^{2}-4 a=4 b^{2}+4 b+1=(2 b+1)^{2}$. Therefore, the equation has two distinct real roots for any $b \neq-\frac{1}{2}$ or $a \neq \frac{1}{4}$.

However, from the given assumptions, it also follows that $a \neq-\frac{1}{4}(b \neq 0)$ and $a \neq-\frac{1}{8}(a \neq b)$.

Conclusion: All pairs $\left(a,-a-\frac{1}{4}\right)$, where $a \in \mathbb{R} \backslash\left\{-\frac{1}{2},-\frac{1}{4},-\frac{1}{8}, 0, \frac{1}{4}\right\}$, satisfy the conditions.

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