# Probability of real Roots of the Equation At² + Bt + C = 0?

Browsing back through Q & A I found this:
http://answers.yahoo.com/question/index;_ylt=Ak69jySnt0dwy9IOIM12Oi0Cxgt.;_ylv=3?qid=20081007064619AAs2Jyy
It reminded me another similar question, asked some time ago (see links below) but I haven't encountered much more interesting generalized question: if A, B, C are...
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Browsing back through Q & A I found this:

http://answers.yahoo.com/question/index;...

It reminded me another similar question, asked some time ago (see links below) but I haven't encountered much more interesting generalized question: if A, B, C are random real coefficients (not necessarily digits!), what is the probability the equation:

At² + Bt + C = 0 to have real roots?

To avoid any misunderstanding what exactly "random real coefficients" should mean, I suggest the following:

Approach 1) Let the point (A,B,C) is uniformly distributed in the cube [-R,R; -R,R; -R,R] /R=const>0/, or, equivalently, A, B, C are uniformly distributed independent random variables in [-R,R] each. Express the probability in terms of R, then let R → ∞.

Approach 2) Same as above, but the point (A,B,C) to be uniformly distributed in a sphere with radius R.

I consider Approach 1) most natural, but feel free to choose 2) instead (that could led to more complicated integrations), or to suggest another eventual approach (and solution) You find reasonable enough. It may be useful to read the following comments and discussions:

http://answers.yahoo.com/question/index;...

http://answers.yahoo.com/question/index;...

and Pascal's excellent explanation:

http://answers.yahoo.com/question/index;_ylt=AjfKrzRmAbT5YiBUfEFiWxDty6IX;_ylv=3?qid=20080314012855AAVWC3v&show=7#profile-info-f70e2c276c18aa87d9308f09887e96b6aa

Of course P(Real Roots) =

= P[(A≠0 and B²-4AC ≥ 0) or (A=0 and B≠0) or (A=0 and B=0 and C=0)]=

= P(B² ≥ 4AC) since P(A=0) = 0. So what is P(B² ≥ 4AC)?

http://answers.yahoo.com/question/index;...

It reminded me another similar question, asked some time ago (see links below) but I haven't encountered much more interesting generalized question: if A, B, C are random real coefficients (not necessarily digits!), what is the probability the equation:

At² + Bt + C = 0 to have real roots?

To avoid any misunderstanding what exactly "random real coefficients" should mean, I suggest the following:

Approach 1) Let the point (A,B,C) is uniformly distributed in the cube [-R,R; -R,R; -R,R] /R=const>0/, or, equivalently, A, B, C are uniformly distributed independent random variables in [-R,R] each. Express the probability in terms of R, then let R → ∞.

Approach 2) Same as above, but the point (A,B,C) to be uniformly distributed in a sphere with radius R.

I consider Approach 1) most natural, but feel free to choose 2) instead (that could led to more complicated integrations), or to suggest another eventual approach (and solution) You find reasonable enough. It may be useful to read the following comments and discussions:

http://answers.yahoo.com/question/index;...

http://answers.yahoo.com/question/index;...

and Pascal's excellent explanation:

http://answers.yahoo.com/question/index;_ylt=AjfKrzRmAbT5YiBUfEFiWxDty6IX;_ylv=3?qid=20080314012855AAVWC3v&show=7#profile-info-f70e2c276c18aa87d9308f09887e96b6aa

Of course P(Real Roots) =

= P[(A≠0 and B²-4AC ≥ 0) or (A=0 and B≠0) or (A=0 and B=0 and C=0)]=

= P(B² ≥ 4AC) since P(A=0) = 0. So what is P(B² ≥ 4AC)?

Update:
Amazing interest so far!
The surface y² = 4xz /I'll use x, y, z instead of A, B, C/ is an elliptic cone with axis through the origin along the vector (1,0,1), intersections, perpendicular to the axis are ellipses, whose axes are in ratio 1 : √2. Here is how it is looking...
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Amazing interest so far!

The surface y² = 4xz /I'll use x, y, z instead of A, B, C/ is an elliptic cone with axis through the origin along the vector (1,0,1), intersections, perpendicular to the axis are ellipses, whose axes are in ratio 1 : √2. Here is how it is looking like:

http://farm4.static.flickr.com/3010/2998...

Using Approach 1) above, I tried to calculate the volume of that part of the cube, that is outside the cone, and indeed got a value, independent of R, but I'd wish to check my result. I hope many of those, who answered the previous similar question, to share their opinions now too - Ksoileau and Scythian - many thanks for answering!

The surface y² = 4xz /I'll use x, y, z instead of A, B, C/ is an elliptic cone with axis through the origin along the vector (1,0,1), intersections, perpendicular to the axis are ellipses, whose axes are in ratio 1 : √2. Here is how it is looking like:

http://farm4.static.flickr.com/3010/2998...

Using Approach 1) above, I tried to calculate the volume of that part of the cube, that is outside the cone, and indeed got a value, independent of R, but I'd wish to check my result. I hope many of those, who answered the previous similar question, to share their opinions now too - Ksoileau and Scythian - many thanks for answering!

Update 2:
Scythian, thanks for the article, I read it very carefully, it emphasizes once more the importance of the precise formulation, what I think I have done above (maybe I shouldn't have used words like "random real", but Approach 1 & 2 make things perfectly clear - we can compare with the well-known...
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Scythian, thanks for the article, I read it very carefully, it emphasizes once more the importance of the precise formulation, what I think I have done above (maybe I shouldn't have used words like "random real", but Approach 1 & 2 make things perfectly clear - we can compare with the well-known Bertrand's Paradox). Now we have 2 distinct problems, shown by Dr D's and JB's excellent contributions (JB's Approach 3 very interesting!), I can confirm their numeric results by computer simulations that I made also.

I have a sound reason to prefer Approach 1 - if (A,B,C) is uniformly distributed in a cube, as described above, the component random variables A,B,C are INDEPENDENT, what is not the case with Approach 2. The volume of that part of the cube, inside the cone is 4 times

∫∫ [D] 2√(xz) dx dz - ∫∫ [D'] (√(xz) - R) dx dz, where

D = { 0 ≤ x ≤ R, 0 ≤ z ≤ R },

D' = { 0 ≤ x ≤ R, 0 ≤ z ≤ R, 4xz ≥ R² }

as shown here:

http://farm4.static.flickr.com/3157/3034...

I have a sound reason to prefer Approach 1 - if (A,B,C) is uniformly distributed in a cube, as described above, the component random variables A,B,C are INDEPENDENT, what is not the case with Approach 2. The volume of that part of the cube, inside the cone is 4 times

∫∫ [D] 2√(xz) dx dz - ∫∫ [D'] (√(xz) - R) dx dz, where

D = { 0 ≤ x ≤ R, 0 ≤ z ≤ R },

D' = { 0 ≤ x ≤ R, 0 ≤ z ≤ R, 4xz ≥ R² }

as shown here:

http://farm4.static.flickr.com/3157/3034...

Update 3:
That's it, Dr D! You got it done!
The answer is shape dependent as JB notes, and if we agree on Approach 1, the rest is somewhat dull integration exercise. Over the weekend I had enough time to check it thoroughly, the analytic result is:
∫∫ [D] 2√(xz) dx dz = 2∫ [0, R]√(x) dx * ∫ [0, R] √z dz = 8R³/9;
∫∫...
show more
That's it, Dr D! You got it done!

The answer is shape dependent as JB notes, and if we agree on Approach 1, the rest is somewhat dull integration exercise. Over the weekend I had enough time to check it thoroughly, the analytic result is:

∫∫ [D] 2√(xz) dx dz = 2∫ [0, R]√(x) dx * ∫ [0, R] √z dz = 8R³/9;

∫∫ [D']((2√(xz) - R) dx dz) = ∫ [x=0 to x=R] (∫ [z=R²/(4x) to z=R] 2√(xz) dz) dx =

= (1/36 + ln 2 /6)R³

The volume V' of the part of the cube inside the cone is:

V' = (31/36 - ln 2 /6)R³

The required probability (according Approach 1):

4V'/(2R)³ = (41 + ln 64)/72 ≈ 0.627206708491 according Windows Calculator.

The answer is shape dependent as JB notes, and if we agree on Approach 1, the rest is somewhat dull integration exercise. Over the weekend I had enough time to check it thoroughly, the analytic result is:

∫∫ [D] 2√(xz) dx dz = 2∫ [0, R]√(x) dx * ∫ [0, R] √z dz = 8R³/9;

∫∫ [D']((2√(xz) - R) dx dz) = ∫ [x=0 to x=R] (∫ [z=R²/(4x) to z=R] 2√(xz) dz) dx =

= (1/36 + ln 2 /6)R³

The volume V' of the part of the cube inside the cone is:

V' = (31/36 - ln 2 /6)R³

The required probability (according Approach 1):

4V'/(2R)³ = (41 + ln 64)/72 ≈ 0.627206708491 according Windows Calculator.

Update 4:
1 - 4V'/(2R)³ on the previous text line, sorry!

Update 5:
And (√(xz) - R) instead of √(xz) only in the inner integral, of course!

Update 6:
We have already a full solution of the problem according Approach 1 and valuable overall comments, so there is no need to keep it open any more.

Try the next, more difficult one:

http://answers.yahoo.com/question/index;...

Try the next, more difficult one:

http://answers.yahoo.com/question/index;...

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