Probability of $x^2+bx+c=0$ having real roots

Question

Consider the quadratic equation $x^2+bx+c=0$, where $b$ and $c$ are $Uni∼[0,1]$. Let $p(b)$ represent the probability that the given equation has a real solution for a fixed value of$ b$.

What is $p(1/2)$?

What is the probability that$ x^2+bx+c=0$ has a real solution ?

$p(1/2)$ = prob $x^2+bx+c=0$ has a real solution)

now $x^2+bx+c=0$ has a real solution if discriminant $≥0$ i.e., $1−16c≥0$ which

is same as $c≤1/16$. since $C∼UNI[0,1]$ the $p(c)≤1/16=1/16$

Is this correct and how to do the second part?

x^2 +1/2x +c=0 can be written as 2x^2 +x+2c= 0 D= 1-42*2c – onelessproblem May 09 '17 at 02:59 — onelessproblem, May 09 '17 at 02:59
First part is correct. Are b and c independent? – John Doe May 09 '17 at 03:00 — John Doe, May 09 '17 at 03:00
Idk, that's not stated in the question. – onelessproblem May 09 '17 at 03:00 — onelessproblem, May 09 '17 at 03:00

score 3 · Accepted Answer · answered May 09 '17 at 03:01

$x^2+bx+c=0$ has a real solution if and only if

$$b^2-4c\le0$$

i.e. $c\le\frac{b^2}{4}$.

Note that when $b\in[0,1]$, $\frac{b^2}{4}\in[0,1]$. For a fixed $b$, the probability that the quadratic equation has a real solution is $\frac{b^2}{4}$.

So if $b,c\sim\text{uni}[0,1]$, the probability that the equation has a real solution is

$$\int_0^1\frac{b^2}{4}db=\frac{1}{12}$$

Probability of $x^2+bx+c=0$ having real roots

1 Answers1

Linked

Related