How many ordered triples of positive integers (a,b,c) with 1<=a,b,c<=5 are there such that ax^2+bx+c has a rational solution?
To find the number of ordered triples (a, b, c) of positive integers with 1 <= a, b, c <= 5 such that ax^2 + bx + c has a rational solution, we can use the Rational Root Theorem.
The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational solution p/q, where p and q are relatively prime integers, then p must divide the constant term of the polynomial (in this case, c), and q must divide the leading coefficient (in this case, a).
Let's analyze the possible values of p and q using this theorem:
1. For p, since p must divide c, we have the following possibilities:
- If c = 1, we can choose p from the set {1}.
- If c = 2, we can choose p from the set {1, 2}.
- If c = 3, we can choose p from the set {1, 3}.
- If c = 4, we can choose p from the set {1, 2, 4}.
- If c = 5, we can choose p from the set {1, 5}.
2. For q, since q must divide a, we have the following possibilities:
- If a = 1, we can choose q from the set {1}.
- If a = 2, we can choose q from the set {1, 2}.
- If a = 3, we can choose q from the set {1, 3}.
- If a = 4, we can choose q from the set {1, 2, 4}.
- If a = 5, we can choose q from the set {1, 5}.
Now, we can combine the possible values of p and q to determine the number of ordered triples (a, b, c) that satisfy the conditions:
1. For each value of c, there are the following possible values of p:
- If c = 1, there is 1 possible value of p.
- If c = 2, there are 2 possible values of p.
- If c = 3, there are 2 possible values of p.
- If c = 4, there are 3 possible values of p.
- If c = 5, there are 2 possible values of p.
2. For each value of a, there are the following possible values of q:
- If a = 1, there is 1 possible value of q.
- If a = 2, there are 2 possible values of q.
- If a = 3, there are 2 possible values of q.
- If a = 4, there are 3 possible values of q.
- If a = 5, there are 2 possible values of q.
To get the final answer, we can multiply the number of possible values of p and q for each case. Then, we sum up the results for all values of c and a:
(1 * 1) + (2 * 2) + (2 * 2) + (3 * 3) + (2 * 2) = 1 + 4 + 4 + 9 + 4 = 22
Therefore, there are 22 ordered triples of positive integers (a, b, c) with 1 <= a, b, c <= 5 such that ax^2 + bx + c has a rational solution.
To find the number of ordered triples of positive integers (a, b, c) with 1 <= a, b, c <= 5 such that ax^2 + bx + c has a rational solution, we can use the discriminant of the quadratic equation.
The discriminant is given by the formula D = b^2 - 4ac. For the quadratic equation to have a rational solution, the discriminant must be a perfect square.
Let's consider the possible values of a, b, and c.
1. If a = 1, then we have:
- b^2 - 4c must be a perfect square.
We can consider all possible values of b and c from 1 to 5 and check if b^2 - 4c is a perfect square.
For each value of b and c, we calculate b^2 - 4c and check if it is a perfect square.
The values of b and c that satisfy this condition are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1), (4, 1), (5, 1).
So, for a = 1, there are 9 possible ordered triples.
2. If a = 2, then we have:
- b^2 - 8c must be a perfect square.
Again, we can consider all possible values of b and c from 1 to 5 and check if b^2 - 8c is a perfect square.
The values of b and c that satisfy this condition are: (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (4, 1), (5, 1).
So, for a = 2, there are 8 possible ordered triples.
3. If a = 3, 4, or 5, then the values of b and c don't matter because the quadratic equation ax^2 + bx + c will always have a rational solution.
Therefore, the total number of ordered triples (a, b, c) such that ax^2 + bx + c has a rational solution is 9 + 8 = 17.