(Roots of Quadratic Equations) A, BB, and CC are selected at random and independently from the interval (0,1)(0,1).

[mario99]

Three numbers $A$, $B$, and $C$ are selected at random and independently from the interval $(0,1)$. Determine the probability that the quadratic equation $Ax^2 + Bx + C = 0$ has real roots. In other words, what fraction of “all possible quadratic equations” with coefficients in $(0,1)$ have real roots?


The solution is

$P(B^2 - 4AC \geq 0) \approx 0.2544134189822131$

Is there an easy way to get this answer?

 

[Steven G]

Where did you get this answer? What topics did you recently learn?

 

[BigBeachBanana]

Three numbers $A$, $B$, and $C$ are selected at random and independently from the interval $(0,1)$. Determine the probability that the quadratic equation $Ax^2 + Bx + C = 0$ has real roots. In other words, what fraction of “all possible quadratic equations” with coefficients in $(0,1)$ have real roots?


The solution is

$P(B^2 - 4AC \geq 0) \approx 0.2544134189822131$

Is there an easy way to get this answer?

Find the CDF of $B^2$ and CDF of $-4AC$, then apply the convolution theorem.

 

[mario99]

Where did you get this answer? What topics did you recently learn?

Thank you.

I got it at the end of the book. I learnt how to find the CDF when the joint density $f$ is known.

Find the CDF of $B^2$ and CDF of $-4AC$, then apply the convolution theorem.

Thank you.

My attempt:

When $a < 0$
$F_{B^2}(a) = 0$

When $a \geq 1$
$F_{B^2}(a) = 1$

When $0 \leq a < 1$
$\displaystyle F_{B^2}(a) = \int_{0}^{a} f(x) \ dx$

How to find the last one when $f(x)$ is unknown?

 

[BigBeachBanana]

Thank you.

My attempt:

When $a < 0$
$F_{B^2}(a) = 0$

When $a \geq 1$
$F_{B^2}(a) = 1$

When $0 \leq a < 1$
$\displaystyle F_{B^2}(a) = \int_{0}^{a} f(x) \ dx$

How to find the last one when $f(x)$ is unknown?

$$\displaystyle F_{B^2}(a) =\Pr(B^2\le a) = \Pr(B\le \sqrt{a})=F_{B}(\sqrt{a})$$

 

[mario99]

$$\displaystyle F_{B^2}(a) =\Pr(B^2\le a) = \Pr(B\le \sqrt{a})=F_{B}(\sqrt{a})$$

Ok. So far so good.

The domain of $F_{B^2}(a)$ is
$a < 0$
or
$0 \leq a < 1$
or
$a > 1$

How to find the domain of $F_{-4AC}(a)$? Is it also similar to the above?

 

[BigBeachBanana]

Ok. So far so good.

The domain of $F_{B^2}(a)$ is
$a < 0$
or
$0 \leq a < 1$
or
$a > 1$

Ok you got the domain, what about the CDF itself?

How to find the domain of $F_{-4AC}(a)$? Is it also similar to the above?

Yes.

 

[mario99]

Ok you got the domain, what about the CDF itself?

$\displaystyle F_{B^2}(a) =\begin{cases}0, & a < 0\\ \sqrt{a}, & 0 \leq a < 1\\1, & a \geq 1\end{cases}$

Yes.

The book says, the domain of $F_{-4AC}$ is
$\displaystyle \begin{cases}a < -4\\-4 \leq a < 0\\a \geq 0\end{cases}$
I am confused!?

 

[BigBeachBanana]

$\displaystyle F_{B^2}(a) =\begin{cases}0, & a < 0\\ \sqrt{a}, & 0 \leq a < 1\\1, & a \geq 1\end{cases}$



The book says, the domain of $F_{-4AC}$ is
$\displaystyle \begin{cases}a < -4\\-4 \leq a < 0\\a \geq 0\end{cases}$
I am confused!?

It seems like you have the book solution. Do you not understand the solution?

Not sure what you're goal here. Find the CDF $F_{AC}(a)$ first then the domain comes after.

 

[mario99]

It seems like you have the book solution. Do you not understand the solution?

Not sure what you're goal here. Find the CDF $F_{AC}(a)$ first then the domain comes after.

The CDF of $F_{B^2-4AC}(a)$ will need the convolution theorem. But the domain of $F_{-4AC}(a)$ will come first.

I don't understand why the domain is like this:

$F_{-4AC}(a) =\begin{cases}0, & a < -4\\ f(a), & -4 \leq a < 0\\1, & a \geq 0\end{cases}$

We will find f(a) later with other techniques. I don't understand how they got this domain!

Note: They should fix their website. It is very slow. It takes 3 days or more to reply to one post!

 

[khansaheb]

Note: They should fix their website. It is very slow. It takes 3 days or more to reply to one post!

Who should fix whose website?
In general, you got a response back within a day, after posting your query!

 

[BigBeachBanana]

The CDF of $F_{B^2-4AC}(a)$ will need the convolution theorem. But the domain of $F_{-4AC}(a)$ will come first.

I don't understand why the domain is like this:

$F_{-4AC}(a) =\begin{cases}0, & a < -4\\ f(a), & -4 \leq a < 0\\1, & a \geq 0\end{cases}$

We will find f(a) later with other techniques. I don't understand how they got this domain!

Note: They should fix their website. It is very slow. It takes 3 days or more to reply to one post!

No. The convolution is between $B^2$ and $AC$, not $A$ and $C$

Follow the same approach as post#5 to find $F_{AC}$ and the domain follows.

 

[mario99]

Who should fix whose website?
In general, you got a response back within a day, after posting your query!

Thank you for fixing the problem. It is working fine now.

No. The convolution is between $B^2$ and $AC$, not $A$ and $C$

Follow the same approach as post#5 to find $F_{AC}$ and the domain follows.

$-4AC \leq a$

$\displaystyle A \geq \frac{a}{-4C}$
Or
$\displaystyle C \geq \frac{a}{-4A}$

I still don't see how that domain was formed.

 

[BigBeachBanana]

Thank you for fixing the problem. It is working fine now.



$-4AC \leq a$

$\displaystyle A \geq \frac{a}{-4C}$
Or
$\displaystyle C \geq \frac{a}{-4A}$

I still don't see how that domain was formed.

Or... $\displaystyle AC \ge -\dfrac{a}{4}$

 

[mario99]

Nice catch, Banana. Please, continue...

 

[BigBeachBanana]

Nice catch, Banana. Please, continue...

No, you continue.

 

[mario99]

No, you continue.

We know that.

$\displaystyle 0 < A < 1$
And
$\displaystyle 0 < C < 1$
Therefore.
$\displaystyle 0 < AC < 1$

When $\displaystyle AC \geq -\frac{a}{4}$, we see that $\displaystyle a = 0$ and $\displaystyle a = -4$ are the critical points.

Is this the reason for the domain to be...?

$\displaystyle a < -4$
$\displaystyle -4 \leq a < 0$
$\displaystyle a \geq 0$

 

[BigBeachBanana]

We know that.

$\displaystyle 0 < A < 1$
And
$\displaystyle 0 < C < 1$
Therefore.
$\displaystyle 0 < AC < 1$

When $\displaystyle AC \geq -\frac{a}{4}$, we see that $\displaystyle a = 0$ and $\displaystyle a = -4$ are the critical points.

Is this the reason for the domain to be...?

$\displaystyle a < -4$
$\displaystyle -4 \leq a < 0$
$\displaystyle a \geq 0$

Yes.

 

[mario99]

Yes.

I think that the fun part of the problem has just begun?

What is your advice for the next step??

 

[BigBeachBanana]

I think that the fun part of the problem has just begun?

What is your advice for the next step??

Re-read post #2.