# Prob of Real Roots

#### maccaman

##### New Member
Hi TalkStats,
This is my first post. If someone could help me solve this problem I would be very grateful.

Question:
A,B,C are independent RV's. Consider the quadratic equation:

A(x^2) + Bx + C = 0.

Find the probability that the given equation has real roots when A,B,C have a U(0,1) distribution?

My work so far:

The probability that the roots are real is the probability that B^2 - 4AC >= 0. i.e. the discriminant is positive.

I have worked out the following from my text.....please see image.

but I am now confused with how to get the solution. Can anyone please show me what to do from here?

Thanks in advance for your help #### vinux

##### Dark Knight
Nice problem.. And you should apply the condition in the range... B2-4AC >=0.
So the ranges could be

A ranges 0 to 1
C ranges 0 to 1
B ranges from -2sqrt(AC) to 2sqrt(AC) => 0 to min(2sqrt(AC) ,1)

#### maccaman

##### New Member

I am lost with the how to apply the condition. Given the pdf of a uniform distribution, what would the actual integration be??

This is where im having trouble

#### vinux

##### Dark Knight
Yes. it seems a difficult problem. Applying the condition is tricky...

Let us split this into multiple sum..
because you have constraint( B2-4AC >=0 ) here and lower and upper bounds ( 0 and 1). So this would make multiple cases.

case1 2sqrt(AC) <=1 and 4C <=1
the above condition we can include in the limit as follows

C ranges 0 to 1/4
A ranges 0 to min(1/(4C),1) => 0 to 1
and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC
= int(C) int(A , 0 to 1 ) 2sqrt(AC) fA fC
= int(C, 0 to 1) {2sqrt(AC) A *2/3}at A = 1 fC
= int(C, 0 to 1) {sqrt(C) *2/3}fC
= {2/3 sqrt(C)C *2/3 }at C=1/4
= 1/18

case2 2sqrt(AC) <=1 and 4C >=1
the above condition we can include in the limit as follows

C ranges 1/4 to 1
A ranges 0 to 1/(4C)
and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC
= int(C) int(A , 0 to 1 ) 2sqrt(AC) fA fC
= int(C, 0 to 1) {2sqrt(AC) A *2/3}at A = 1/4C fC
= int(C, 0 to 1) {1/4C *2/3}fC
= {1/6 log(C) }at C=1 - {1/6 log(C) }at C=1/4

case3 2sqrt(AC) >=1 and C<=1/4
the above condition we can include in the limit as follows

C ranges 0 to 1/4
A ranges 1/4C to 1 => 1 to 1
and B ranges 0 to 2sqrt(AC)

area =0

case4 2sqrt(AC) >=1 and C >=1/4
the above condition we can include in the limit as follows

C ranges 1/4 to 1
A ranges 1/(4C) to 1
and B ranges 0 to 1
calculate area..

You can sum the four cases and get the proabability.

Please cross check the integral, because i haven't done integration for long time
please let me know if you want to have clarification on any steps apart from integration.