What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots? A number $a$ is chosen at random within the interval $(-1, 1)$. What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots?
For it to have its real roots, we must guarantee that $1-4a \geq 0$, or $a\leq \frac{1}{4}$.
It is no longer clear to me what I have to do.
 A: We want the probability that $a\in (-1,\frac{1}{4}]$ given that it is uniformly chosen from the interval $(-1,1)$.
Since the interval $(-1,\frac{1}{4})$ has length $\frac{5}{4}$ and the interval $(-1,1)$ has length $2$, the probability is
$$\frac{\frac{5}{4}}{2}$$
$$\boxed{\frac{5}{8}}$$
A: This equations have two distinct real solutions iff $a\in(-1,0)\cup (0,\frac{1}{4})$. (When $a\in\{0,\frac{1}{4}\}$ it has one real solution) Therefore the probability is
$$P=\frac{l((-1,0)\cup (0,\frac{1}{4}))}{l(-1,1)}$$
Here $l(I)$ is the length of interval $I$.
$$P=\frac{1+\frac{1}{4}}{2}=\frac{5}{8}$$
A: $$a \in(-1,1) \land a \leq \frac{1}{4} \iff a \in(-1,\frac{1}{4}]$$
because $(-1,1) \cup (-\infty,\frac{1}{4}] = (-1,\frac{1}{4}]$
We can therefore, divide the length of both intervals to get the probability:
$$P(a \in(-1,\frac{1}{4}]) = \frac{|\frac{1}{4} - \left(-1\right)|}{|1 - \left(-1\right)|} = \frac{\frac{5}{4}}{2} = \frac{5}{8}$$
The fact that $(-1,\frac{1}{4}]$ doesn't include $-1$ and $\left(-1,1\right)$ doesn't include either of its endpoints doesn't make a difference because they are single points that can be considered insignificant compared to the infinite number of reals in the intervals.
