The coefficients a, b, c in the quadratic equation a x²+bx+c=0 are from the set {1,2,3,4,5,6}. If the probability of this equation having one real root bigger than the other is p, then 216p equals :

Solution

Maths sir · 4 steps

One root bigger than the other means two different real roots, so count triples with b² > 4ac.

1. Translate the condition. The roots are real and unequal exactly when the discriminant is positive, so we need b² > 4ac. Each of a, b, c has 6 choices, so there are 6³ = 216 equally likely equations.
2. Fix b first. For each value of b, count the ordered pairs (a, c) with ac < b²/4. For b = 1 and b = 2 we would need ac < 1, which is impossible, so they give nothing.

   b	Need	Ordered pairs (a, c)	Count
   3	ac ≤ 2	(1,1), (1,2), (2,1)	3
   4	ac ≤ 3	(1,1), (1,2), (2,1), (1,3), (3,1)	5
   5	ac ≤ 6	products 1, 2, 3, 4, 5, 6 give 1+2+2+3+2+4 pairs	14
   6	ac ≤ 8	the 14 above, plus (2,4), (4,2) for product 8	16

3. Add up. Favourable triples: 3 + 5 + 14 + 16 = 38. Product 7 adds nothing, because 7 cannot be written as a product of two numbers from the set.
4. So. p = 38/216, so 216p = 38, which is option (a).

The idea to keep. When coefficients come from a small set, fix the middle coefficient b and count ordered pairs (a, c) under the discriminant condition. Remember that (1,2) and (2,1) are different equations.

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