The existence of distinct rational roots means that the given quadratic splits into linear factors. Then, since $p_{1}, p_{3}$ are both prime, we get that the following are the only possible factorizations: - $(p_{1} x-p_{3})(x+1) \Rightarrow p_{2}=p_{1}-p_{3}$ - $(p_{1} x+p_{3})(x-1) \Rightarrow p_{2}=-p_{1}+p_{3}$ - $(p_{1} x-1)(x+p_{3}) \Rightarrow p_{2}=p_{1} p_{3}-1$ - $(p_{1} x+1)(x-p_{3}) \Rightarrow p_{2}=-p_{1} p_{3}+1$ In the first case, observe that since $p_{2}+p_{3}=p_{1}$, we have $p_{1}>2$, so $p_{1}$ is odd and exactly one of $p_{2}, p_{3}$ is equal to 2. Thus, we get a solutions for every pair of twin primes below 50, which we enumerate to be $(3,5),(5,7),(11,13),(17,19),(29,31),(41,43)$, giving 12 solutions in total. Similarly, the second case gives $p_{1}+p_{2}=p_{3}$, for another 12 solutions. In the third case, if $p_{1}, p_{3}$ are both odd, then $p_{2}$ is even and thus equal to 2. However, this gives $p_{1} p_{3}=3$, which is impossible. Therefore, at least one of $p_{1}, p_{3}$ is equal to 2. If $p_{1}=2$, we get $p_{2}=2 p_{3}-1$, which we find has 4 solutions: $(p_{2}, p_{3})=(3,2),(5,3),(13,7),(37,19)$. Similarly, there are four solutions with $p_{3}=2$. However, we count the solution $(p_{1}, p_{2}, p_{3})=(2,3,2)$ twice, so we have a total of 7 solutions in this case. Finally, in the last case $p_{2}=-p_{1} p_{3}+1<-(2)(2)+1<0$ so there are no solutions. Hence, we have a total of $12+12+7=31$ solutions.