a, b, c, A and C are real numbers that satisfy the equation
(ax2+bx+c)2=Ax4+16x3+28x2+24x+C
How many possible ordered triples of (a,b,c) are there?
To determine the number of possible ordered triples of (a,b,c), we need to find the values of a, b, and c that satisfy the given equation.
Let's start by expanding the equation on both sides:
(a^2x^4 + 2abx^3 + 2acx^2 + b^2x^2 + 2bcx + c^2) = Ax^4 + 16x^3 + 28x^2 + 24x + C
By comparing the coefficients on both sides, we have the following equations:
1) a^2 = A
2) 2ab = 0
3) 2ac + b^2 = 0
4) 2bc = 16
5) c^2 = C
1) Since a^2 = A, we can say that a = sqrt(A) or a = -sqrt(A). Therefore, there are 2 possible values for a.
2) Since 2ab = 0, this implies that either a = 0 or b = 0. If a = 0, then the equation in 3) becomes 2ac + b^2 = 0. If b = 0, then the equation in 4) becomes 2ac = 0. In both cases, we have only one free variable, c, and we can choose any real number for c. Therefore, there is an infinite number of possible values for a, b, and c.
Combining all the possibilities, the number of possible ordered triples of (a,b,c) can be summarized as follows:
If a ≠ 0 and b ≠ 0: 2 ordered triples (a, b, c) satisfying the given equation.
If a = 0 and b ≠ 0: Infinite ordered triples (0, b, c) satisfying the given equation.
If a ≠ 0 and b = 0: Infinite ordered triples (a, 0, c) satisfying the given equation.
If a = 0 and b = 0: Infinite ordered triples (0, 0, c) satisfying the given equation.
In conclusion, the number of possible ordered triples of (a,b,c) can vary depending on whether a and/or b equal zero, resulting in either 2 or an infinite number of solutions.