How many ordered triples of positive integers (a,b,c) with 1¡Üa,b,c¡Ü5 are there such that ax^2+bx+c has a rational solution?
To find the number of ordered triples (a, b, c) with positive integers satisfying 1≤a,b,c≤5 such that the quadratic equation ax^2+bx+c has a rational solution, we can go through each possible combination of values for a, b, and c and check if the discriminant of the quadratic equation is a perfect square.
The discriminant, denoted as Δ, of the quadratic equation ax^2+bx+c is given by Δ = b^2 - 4ac.
1. For each value of a from 1 to 5:
- For each value of b from 1 to 5:
- For each value of c from 1 to 5:
- Calculate the discriminant Δ = b^2 - 4ac.
- Check if the discriminant is a perfect square.
If the discriminant is a perfect square, increment a counter variable as we found a valid triple.
Once we finish going through all combinations of a, b, and c, the value of the counter variable will give us the total number of ordered triples (a, b, c) satisfying the given conditions.
It's essential to note that the quadratic equation ax^2+bx+c may have more than one rational solution, but we only need to count the distinct triples (a, b, c) that meet the given criteria.