What is the probability that the root of ax^2+bx+c=0 are real numbers?

if a, b & c are positive digits, then there are 9 possible values for all of these each .

thus

(a,b,c) has 9^3 possible combinations ....

we need to find such a combination which will yield a positive discriminant ...

b^2 - 4ac ≥ 0

b^2 ≥ 4ac

[i will just do this manually, i dont know any approach that will make a simple combination to solve this ....]

if b = 1

no possible values for a & c

if b = 2

a = c = 1 ... thus (1,2,1) only

if b = 3

(1,3,1) , (1,3,2) , (2,3,1)

if b = 4

(1,4,1) , (1,4,2) , (1,4,3) , (1,4,4)

(2,4,1) , (2,4,2)

(3,4,1) & (4,4,1)

if b = 5

(1,5,1) ... (1,5,6)

(2,5,1) ... (2,5,3)

(3,5,1) , (3,5,2)

(4,5,1) & (5,5,1) & (6,5,1)

if b = 6

(1,6,1) ... (1,6,9)

(2,6,1) ... (2,6,4)

(3,6,1) ... (3,6,3)

(4,6,1) , (4,6,2)

(5,6,1) ... (9,6,1)

if b = 7

(1,7,1) ... (1,7,9)

(2,7,1) ... (2,7,6)

(3,7,1) ... (3,7,4)

(4,7,1) ... (4,7,3)

(5,7,1) , (5,7,2)

(6,7,1) , (6,7,2)

(7,7,1) .. (9,7,1)

if b = 8

(1,8,1) ... (1,8,9)

(2,8,1) ... (2,8,8)

(3,8,1) ... (3,8,5)

(4,8,1) ... (4,8,4)

(5,8,1) ... (5,8,3)

(6,8,1) , (6,8,2)

(7,8,1) , (7,8,2)

(8,8,1) , (8,8,2)

(9,8,1)

finally if b = 9

(1,9,1) ... (1,9,9)

(2,9,1) ... (2,9,9)

(3,9,1) ... (3,9,6)

(4,9,1) ... (4,9,5)

(5,9,1) ... (5,9,4)

(6,9,1) ... (6,9,3)

(7,9,1) , (7,9,2)

(8,9,1) , (8,9,2)

(9,9,1) , (9,9,2)

thus

1 +

3 +

4 + 2 + 2 +

6 + 3 + 2 + 3 +

9 + 4 + 3 + 2 + 5 +

9 + 6 + 4 + 3 + 2 + 2 + 3 +

9 + 8 + 5 + 4 + 3 + 2 + 2 + 2 + 1 +

9 + 9 + 6 + 5 + 4 + 3 + 2 + 2 + 2

= 156

thus the probability is 156 / 729 .

i assume your shorthand is from possibility thought and is asserting the variables are are self sustaining and identically allotted uniformly contained in the variety 0 to a million. in spite of this, perchance no longer. besides it somewhat is the possibility that B^2 > 4AC . by way of fact the values are decrease than a million, B^2 will constantly be smaller than B and is for this reason probable to be smaller than 4AC provided that the variables are self sustaining and the two allotted. no longer lots probability of actual roots.

If b^2 - 4ac>0,it has two roots,

if b^2 - 4ac=0 ,it has two equal roots

if b^2-4ac<0,it has two imaginary roots