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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.

Gabriela
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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 $$\frac{\frac{5}{4}}{2}$$ $$\boxed{\frac{5}{8}}$$

Alan Abraham
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  • You might also mention that $a = 0$ is disallowed, as well as $a = (1/4)$, since the question asks for the probability of two real roots. – user2661923 Aug 06 '21 at 04:53
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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 $I$. $$P=\frac{1+\frac{1}{4}}{2}=\frac{5}{8}$$

Sahan Manodya
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    The probability of having real roots for equation is same as the probability of having two real distinct roots because the set ${a\in(-1,1) : ax^2+x+1=0 \text{ has only one root}}$ is finite. – Sahan Manodya Aug 06 '21 at 03:41
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$$a \in(-1,1) \land a \leq \frac{1}{4} \iff a \in(-1,\frac{1}{4}]$$

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.