\[ax^2+bx+c=0\]
will have real solutions if \[b^2-4ac \ge 0\]
. What is the probability that a randomly generated quadratic equation, with \[a, \: b, \: c\]
all positive one digit numbers, will have solutions? All possible combinations of \[a, \: b, \: c\]
returning real solutions are given in the table.\[b\] |
\[(a,c)\] |
Number of possibilities |
| 0 | \[\begin{equation} \begin{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} \end{equation}\] |
9 |
| 1 | \[\begin{equation} \begin{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} \end{equation}\] |
9 |
| 2 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \:(9,0), \\&(1,1)\end{aligned} \end{equation}\] |
10 |
| 3 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1)\end{aligned} \end{equation}\] |
12 |
| 4 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (3,1), \: (4,1), \: (2,2)\end{aligned} \end{equation}\] |
17 | 5 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(3,1), \: (4,1), \\&(5,1), \: (6,1), \: (2,2), \: (2,3), \: (3,2)\end{aligned} \end{equation}\] |
23 |
| 6 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \:(3,2), (4,2), \: (3,3)\end{aligned} \end{equation}\] |
32 |
| 7 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (3,2), (4,2), \: (5,2), \: (6,2), \: (3,3), \\&(3,4), \: (4,3)\end{aligned} \end{equation}\] |
38 |
| 8 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (2,7), \: (2,8), \: (3,2), \: (4,2), \: (5,2), \\&(6,2), \: (7,2), \: (8,2), \: (3,3), \: (3,4), \: (3,5), \: (4,3), \: (5,3), \: (4,4)\end{aligned} \end{equation}\] |
44 |
| 9 | \[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (2,7), \: (2,8), \: (2,9), \: (3,2), \: (4,2), \\&(5,2), \: (6,2), \: (7,2), \: (8,2), \: (9,2), \:(3,3), \: (3,4), \: (3,5), \: (3,6), \\&(4,3), \: (5,3), \: (6,3), \: (4,4), \: (4,5), \: (5,4)\end{aligned} \end{equation}\] |
51 |
\[a\]
(the digits 1 to 9) and 10 possibilities each for \[b\]
and \[c\]
so 900 possible values altogether. The probability is \[\frac{245}{900}=\frac{49}{180}\]
.