## Probabilitiy of a Randomly Generated Quadrtic Having Solutions

A quadratic equation
$ax^2+bx+c=0$
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{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} 9 1 \begin{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} 9 2 \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \:(9,0), \\&(1,1)\end{aligned} 10 3 \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} 12 4 \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} 17 5 \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} 23 6 \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} 32 7 \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} 38 8 \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} 44 9 \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} 51
There are 3245 combinations. There are 9 possible values for
$a$
(the digits 1 to 9) and 10 possibilities each for
$b$
and
$c$
so 900 possible values altogether. The probability is
$\frac{245}{900}=\frac{49}{180}$
.

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