Let a, b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1,2,3,4. If the probability that a x²+b x+c=0 has all real roots is m/n, gcd(m, n)=1, then m+n is equal to _________.

Solution

Maths sir · 3 steps

Fix b first, since it is the only one squared, then count the (a, c) pairs that keep b² ≥ 4ac.

1. Condition for real roots. The roots are real when the discriminant is not negative, that is b² ≥ 4ac, or ac ≤ b²/4. Equal roots are also real, so the equality case counts. There are 4³ = 64 equally likely outcomes in all.

   b	Need ac ≤	Pairs (a, c) that work	Count
   1	0.25	none, since ac ≥ 1	0
   2	1	(1,1)	1
   3	2.25	(1,1), (1,2), (2,1)	3
   4	4	ac = 1: 1 pair; ac = 2: 2 pairs; ac = 3: 2 pairs; ac = 4: (1,4), (2,2), (4,1)	8

2. Add up. Favourable outcomes = 0 + 1 + 3 + 8 = 12. So the probability is 12/64 = 3/16, already in lowest terms.
3. So. m = 3 and n = 16, which gives m + n = 19.

The common slip: 9. Using b² > 4ac drops the equal-root cases (1,1) for b = 2 and (1,4), (2,2), (4,1) for b = 4. That leaves 8/64 = 1/8 and gives 9.

The idea to keep. For discriminant counting with small dice values, fix the squared coefficient and list the products ac that fit. Remember that real roots means ≥ 0, so equal roots count too.

Answer from the NTA final answer key. Solution by PrepWiser.