For how many ordered triples of positive real numbers (a,b,c), do the equations ax^2+2bx+c, bx^2+2cx+a and cx^2+2ax+b each have 2 distinct real roots?
To find the number of ordered triples of positive real numbers (a, b, c) for which each of the given equations has two distinct real roots, we can use the discriminant.
The discriminant of a quadratic equation of the form ax^2+bx+c is given by b^2 - 4ac.
For the first equation, ax^2+2bx+c, to have two distinct real roots, the discriminant must be greater than zero. Thus, we have b^2 - 4ac > 0.
For the second equation, bx^2+2cx+a, to have two distinct real roots, the discriminant must also be greater than zero. Therefore, we have (2c)^2 - 4b(a) > 0.
For the third equation, cx^2+2ax+b, to have two distinct real roots, the discriminant should be greater than zero. Hence, we have (2a)^2 - 4c(b) > 0.
Simplifying the above inequalities, we have:
b^2 - 4ac > 0 -----(1)
4c^2 - 4ab > 0 -----(2)
4a^2 - 4bc > 0 -----(3)
Dividing eq. (1) by 4, eq. (2) by 4a, and eq. (3) by 4b, we get:
b/2 > √(ac) -----(4)
c > √(ab) -----(5)
a > √(bc) -----(6)
From the above inequalities, we can observe that a, b, and c must all be positive numbers, as they are square roots, and we are considering positive real numbers.
Now, let's analyze the conditions for (a, b, c) for which the given inequalities hold:
From eq. (4), we can deduce that b/2 > √(ac), which implies b^2 > 4ac.
This can be satisfied when b^2 = 4ac + k, for some positive constant k.
Similarly, from eq. (5), we have c > √(ab), which implies c^2 > ab.
This can be satisfied when c^2 = ab + m, for some positive constant m.
Finally, from eq. (6), we get a > √(bc), which implies a^2 > bc.
This can be satisfied when a^2 = bc + n, for some positive constant n.
Hence, the positive real numbers (a, b, c) must satisfy a^2 = bc + n, b^2 = 4ac + k, and c^2 = ab + m, where n, k, m > 0.
Now, to determine the number of ordered triples (a, b, c), we can use a combinatorial approach.
Suppose we express a^2 = bc + n as a^2 - bc = n.
Since a, b, and c are positive real numbers, we observe that for each value of a, we have a unique pair (b, c) that satisfies the equation. Thus, the number of ways to choose (b, c) is equal to the number of ordered pairs (b, c). This is equivalent to the number of positive divisors of a^2 - n.
Similarly, we can obtain the number of ordered pairs (a, b) and (a, c) for the equations b^2 = 4ac + k and c^2 = ab + m, respectively, using the same reasoning.
Therefore, the total number of ordered triples (a, b, c) is equal to the product of the number of positive divisors for a^2 - n, b^2 - k, and c^2 - m.
In summary, the number of ordered triples of positive real numbers (a, b, c) for which the equations ax^2 + 2bx + c, bx^2 + 2cx + a, and cx^2 + 2ax + b each have two distinct real roots is given by the product of the number of positive divisors for the expressions a^2 - n, b^2 - k, and c^2 - m.