Let us consider the quadratic equation: ax˛ + bx + c = 0.

We assign values to the coefficients a, b and c by throwing a die.

What is the probability that the equation will have real roots ?

The probability is a little less than 1/5.

The quadratic equation states that if
ax²+bx+c=0,
then
x=[-b±√(b²-4ac)]/2a.

The part under the radical, b²-4ac, is called the determinant, because it determines what types of roots the equation will have. If the determinant is zero, then that term drops out, leaving one real root to the equation. If it is positive, there are two real roots, and if it is negative there are two imaginary or complex roots.

I will assume for this problem, since it's not stated, that the condition of 'real roots' includes a single real root, and we are looking for the probability that b²-4ac will be nonnegative.

Another way of saying that b²-4ac is nonnegative, is saying that:
b²≥4ac .
Further:
¼b²≥ac .

Let's say we roll die B first. There are six possible values, of course. There are 6×6=36 possible ways to roll ac, and if we determine ¼b², this must be greater than or equal to ac.

If b is 1, then ¼b²=¼, and it is impossible for ac to be less than this. Therefore there is a probability of zero that the equation will have real roots if b=1.

If b=2, then ¼b²=1. The only way the could form a valid equation is if a and c are both 1, leaving -1 as the only root to the equation. Therefore, if b is 2, there is a 1/36 chance that the equation will have real roots.

If b=3, then ¼b²=9/4=2¼. There are three ordered pairs (a, c) for which this will work: (1, 1), (1, 2), and (2, 1). The probability of any of these is 3/36=1/12.

If b=4, ¼b²=4, and the valid values of (a, c) are:
(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1), (4, 1). The chance of any of these is 8/36=2/9.

If b=5, ¼b²=25/4=6¼, and (a, c) must be one of:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (5, 1), (6, 1). 14 of the 36 combinations are valid, so the chance of getting one of them is 14/36=7/18.

Finally, if b is 6, ¼b²=9, and (a, c) can be any of:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1), (6, 1). These are 17 out of 36 values.

In sum, the chance of b showing up as any given value is, of course, 1/6. The overall probability of finding any of the permutations described here is 1/6 of their sum, or: 1/6(0+1/36+1/12+2/9+7/36+17/36), or 43/(36×6). This evaluates to 0.199074074074..., which is the probability we are looking for.

Another approach could be to list all the valid ordered sets (a, b, c) of of 6×6×6=216 possible values. There will be 43, of course, yielding the same probability as above. The 43 valid ordered sets are:(1, 2, 1), (1, 3, 1), (1, 3, 2), (1, 4, 1), (1, 4, 2), (1, 4, 3), (1, 4, 4), (1, 5, 1), (1, 5, 2), (1, 5, 3), (1, 5, 4), (1, 5, 5), (1, 5, 6), (1, 6, 1), (1, 6, 2), (1, 6, 3), (1, 6, 4), (1, 6, 5), (1, 6, 6), (2, 3, 1), (2, 4, 1), (2, 4, 2), (2, 5, 1), (2, 5, 2), (2, 5, 3), (2, 6, 1), (2, 6, 2), (2, 6, 3), (2, 6, 4), (3, 4, 1), (3, 5, 1), (3, 5, 2), (3, 6, 1), (3, 6, 2), (3, 6, 3), (4, 4, 1), (4, 5, 1), (4, 6, 1), (4, 6, 2), (5, 5, 1), (5, 6, 1), (6, 5, 1), (6, 6, 1). As we can see, there are indeed 43 sets. Any other values will produce a negative determinant, so our solution is complete.

Posted by DJ on 2003-08-20 10:14:16