Suppose $b$ a random number in the interval $(-3,3)$ Which the probability of the equation $x^2+bx+1=0$ have a least a real root? 
Suppose $b$ a random number in the interval $(-3,3)$ Which the probability of the equation $x^2+bx+1=0$ have a least a real root?

I don't know how to start this exercise. Can someone help me?
 A: We need the discriminant $\Delta$ to be $\geq 0$. For a quadratic $ax^2+bx+c$, the discriminant is $b^2-4ac$. In our case, the discriminant is $b^2-4$. Note that 
$$b^2-4 \geq 0 \implies b \in (-3,-2] \cup [2,3)$$
which has measure $2$. Thus, the probability is $\dfrac{2}{6} = \dfrac{1}{3}$.
A: The equation $x^2+bx+1=0$ will have at least one real root if and only if the discriminant is $D=b^2-4ac$ is non-negative which means $D\ge0, b^2-4ac\ge0$.
So, by solving for discriminant, we get $b^2\ge4$ and $|b|\ge2$. So the favourable outcomes are on the interval $(-3,-2]$ and $[2,3)$ which has a combined length of $2$ and all the  outcomes on $(3,3)$ has the length of $6$, so the probability is $\dfrac26=\dfrac13$
A: Hint: The polynomial $f(x) = x^2+bx+1$ has a real root if there is an $x$ s.t. $f(x) \le 0$ is satisfied [make sure you can see why].
The minimum of $f(x)$ is achieved when $df(x)/dx = 2x+b$ is 0, at $x=\frac{-b}{2}$.
The values of $b \in (-3,3)$ s.t. the inequality $f(-\frac{b}{2}) = \frac{b^2}{4} - \frac{b^2}{2}+1 =-\frac{b^2}{4}+1 \le 0$ is satisfied
are $|b| \ge 2$.
Note that $|b| \ge 2$ holds for precisely $\frac{1}{3}$ of $b \in (-3,3)$, so the probability is $\frac{1}{3}$.
