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

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mario99

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


The solution is

[imath]P(B^2 - 4AC \geq 0) \approx 0.2544134189822131[/imath]

Is there an easy way to get this answer?
 
mario99 said:
Three numbers [imath]A[/imath], [imath]B[/imath], and [imath]C[/imath] are selected at random and independently from the interval [imath](0,1)[/imath]. Determine the probability that the quadratic equation [imath]Ax^2 + Bx + C = 0[/imath] has real roots. In other words, what fraction of “all possible quadratic equations” with coefficients in [imath](0,1)[/imath] have real roots?


The solution is

[imath]P(B^2 - 4AC \geq 0) \approx 0.2544134189822131[/imath]

Is there an easy way to get this answer?
Click to expand...
Find the CDF of [imath]B^2[/imath] and CDF of [imath]-4AC[/imath], then apply the convolution theorem.
 
Steven G said:
Where did you get this answer? What topics did you recently learn?
Click to expand...
Thank you.

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



BigBeachBanana said:
Find the CDF of [imath]B^2[/imath] and CDF of [imath]-4AC[/imath], then apply the convolution theorem.
Click to expand...
Thank you.

My attempt:

When [imath]a < 0[/imath]
[imath]F_{B^2}(a) = 0[/imath]

When [imath]a \geq 1[/imath]
[imath]F_{B^2}(a) = 1[/imath]

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

How to find the last one when [imath]f(x)[/imath] is unknown?
 
mario99 said:
Thank you.

My attempt:

When [imath]a < 0[/imath]
[imath]F_{B^2}(a) = 0[/imath]

When [imath]a \geq 1[/imath]
[imath]F_{B^2}(a) = 1[/imath]

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

How to find the last one when [imath]f(x)[/imath] is unknown?
Click to expand...
[math]\displaystyle F_{B^2}(a) =\Pr(B^2\le a) = \Pr(B\le \sqrt{a})=F_{B}(\sqrt{a})[/math]
 
BigBeachBanana said:
[math]\displaystyle F_{B^2}(a) =\Pr(B^2\le a) = \Pr(B\le \sqrt{a})=F_{B}(\sqrt{a})[/math]
Click to expand...
Ok. So far so good.

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

How to find the domain of [imath]F_{-4AC}(a)[/imath]? Is it also similar to the above?
 
mario99 said:
[imath]\displaystyle F_{B^2}(a) =\begin{cases}0, & a < 0\\ \sqrt{a}, & 0 \leq a < 1\\1, & a \geq 1\end{cases} [/imath]



The book says, the domain of [imath]F_{-4AC}[/imath] is
[imath]\displaystyle \begin{cases}a < -4\\-4 \leq a < 0\\a \geq 0\end{cases} [/imath]
I am confused!?
Click to expand...
It seems like you have the book solution. Do you not understand the solution?

Not sure what you're goal here. Find the CDF [imath]F_{AC}(a)[/imath] first then the domain comes after.
 
BigBeachBanana said:
It seems like you have the book solution. Do you not understand the solution?

Not sure what you're goal here. Find the CDF [imath]F_{AC}(a)[/imath] first then the domain comes after.
Click to expand...
The CDF of [imath]F_{B^2-4AC}(a)[/imath] will need the convolution theorem. But the domain of [imath]F_{-4AC}(a)[/imath] will come first.

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

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

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!
 
mario99 said:
Note: They should fix their website. It is very slow. It takes 3 days or more to reply to one post!
Click to expand...
Who should fix whose website?
In general, you got a response back within a day, after posting your query!
 
Last edited:
mario99 said:
The CDF of [imath]F_{B^2-4AC}(a)[/imath] will need the convolution theorem. But the domain of [imath]F_{-4AC}(a)[/imath] will come first.

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

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

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!
Click to expand...
No. The convolution is between [imath]B^2[/imath] and [imath]AC[/imath], not [imath]A[/imath] and [imath]C[/imath]

Follow the same approach as post#5 to find [imath]F_{AC}[/imath] and the domain follows.
 
khansaheb said:
Who should fix whose website?
In general, you got a response back within a day, after posting your query!
Click to expand...
Thank you for fixing the problem. It is working fine now.


BigBeachBanana said:
No. The convolution is between [imath]B^2[/imath] and [imath]AC[/imath], not [imath]A[/imath] and [imath]C[/imath]

Follow the same approach as post#5 to find [imath]F_{AC}[/imath] and the domain follows.
Click to expand...
[imath]-4AC \leq a[/imath]

[imath]\displaystyle A \geq \frac{a}{-4C}[/imath]
Or
[imath]\displaystyle C \geq \frac{a}{-4A}[/imath]

I still don't see how that domain was formed.
 
mario99 said:
Thank you for fixing the problem. It is working fine now.



[imath]-4AC \leq a[/imath]

[imath]\displaystyle A \geq \frac{a}{-4C}[/imath]
Or
[imath]\displaystyle C \geq \frac{a}{-4A}[/imath]

I still don't see how that domain was formed.
Click to expand...
Or... \(\displaystyle AC \ge -\dfrac{a}{4}\)
 
BigBeachBanana said:
No, you continue.
Click to expand...
We know that.

[imath]\displaystyle 0 < A < 1[/imath]
And
[imath]\displaystyle 0 < C < 1[/imath]
Therefore.
[imath]\displaystyle 0 < AC < 1[/imath]

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

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

[imath]\displaystyle a < -4[/imath]
[imath]\displaystyle -4 \leq a < 0[/imath]
[imath]\displaystyle a \geq 0[/imath]
 
mario99 said:
We know that.

[imath]\displaystyle 0 < A < 1[/imath]
And
[imath]\displaystyle 0 < C < 1[/imath]
Therefore.
[imath]\displaystyle 0 < AC < 1[/imath]

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

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

[imath]\displaystyle a < -4[/imath]
[imath]\displaystyle -4 \leq a < 0[/imath]
[imath]\displaystyle a \geq 0[/imath]
Click to expand...
Yes.
 
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