# Find probability that the equation $x^2+Bx+C=0$ has 2 distinct roots.

Let $$B,C$$ independent random variables such that $$B\sim \operatorname{exp}(\lambda),C\sim U[0,1]$$.

I have 2 questions about the solution:

1. "We're looking for the probability that $$\mathbb{P}(4B^2-4C>0)$$". Why does the coefficient of $$B^2$$ is $$4$$ and not $$1$$? Maybe it's a mistake?
2. Why does this equality hold? $$\\ f_{B^2,C}(t,s)=f_{B^2|C}(t|s) \$$
• 1 is a mistake, no question. – Matt Samuel Jan 1 at 17:19

If $$s \in [0,1]$$, we have $$f_C(s)=1$$ since $$C$$ follows $$U[0,1]$$.
Hence, when $$s \in [0,1]$$, $$f_{B^2, C}(t,s) = f_{B^2|C}(t|S)f_C(s)=f_{B^2|C}(t|S)$$