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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?

Key Flex
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rcoder
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    There is a simple test to tell whether a given quadratic equation has real roots. Do you know it? – Arthur Jun 27 '18 at 22:04
  • A more precise way to set up your problem would be to describe the probability distribution of $B$ in $(-3,3)$, eg. a uniform distribution. Note that it is customary to use capital letters for random variable to make the distinction between randomly distributed values and variables with fixed (if unknown) values. – hardmath Jun 27 '18 at 22:09
  • have a least a real root Same probability as having both roots real. – dxiv Jun 27 '18 at 22:09
  • Perhaps you will be interested in this related Q & A. You need $P(b^2 - 4 > 0)$ or $P(b < -2) + P(b > 2) = 1/3$ – BruceET Jun 27 '18 at 22:23

3 Answers3

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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 $\dfrac{2}{6} = \dfrac{1}{3}$.

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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 $\dfrac26=\dfrac13$

Key Flex
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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 $\frac{1}{3}$.

Mike
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