Consider the following system of inequalities:
(c−1)x^2+2cx+c+4≤0 ---- (1)
cx^2+2(c+1)x+(c+1)≥0 ---- (2)
The sum of all real values of c, such that the system has a unique solution, can be written as a/b, where a and b are coprime positive integers. What is the value of a+b?
Details and assumptions
c can be negative.
The system has a unique solution if there is only 1 real value x which is satisfied throughout.
To find the values of c for which the system of inequalities has a unique solution, we need to examine the roots of the quadratic equations.
Let's start by analyzing equation (1):
(c−1)x^2+2cx+c+4≤0
The roots of this quadratic equation are given by the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = (c-1), b = 2c, and c = (c+4).
To have a unique solution, this quadratic equation must have either one real root or a repeated real root (meaning the discriminant b^2 - 4ac is equal to zero).
Now, let's consider equation (2):
cx^2+2(c+1)x+(c+1)≥0
Similarly, the roots of this quadratic equation are given by the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = c, b = 2(c+1), and c = (c+1).
To have a unique solution, this quadratic equation must have either one real root or a repeated real root.
To find the values of c for which the system has a unique solution, we need to analyze the conditions under which both equations (1) and (2) have either one real root or a repeated real root.
Let's consider each case separately:
Case 1: Both equations have one real root:
For both equations (1) and (2) to have one real root, their discriminants must be equal to zero.
For equation (1):
b^2 - 4ac = (2c)^2 - 4(c-1)(c+4) = 0
Simplifying this equation, we get:
c^2 - 10c + 16 = 0
Solving this quadratic equation, we find two real roots: c = 2 and c = 8.
For equation (2):
b^2 - 4ac = (2(c+1))^2 - 4c(c+1) = 0
Simplifying this equation, we get:
c^2 + 2c - 4 = 0
Solving this quadratic equation, we find one real root: c = -2.
Thus, for these two equations to have a unique solution, the values of c can be 2, 8, and -2.
Now let's move on to the next case:
Case 2: Both equations have a repeated real root:
For both equations (1) and (2) to have a repeated real root, their discriminants must be equal to zero.
For equation (1):
b^2 - 4ac = (2c)^2 - 4(c-1)(c+4) = 0
Using the quadratic formula to simplify and solve this equation, we find:
c = -2/3
For equation (2):
b^2 - 4ac = (2(c+1))^2 - 4c(c+1) = 0
Using the quadratic formula to simplify and solve this equation, we find:
c = -1/3
Therefore, for these two equations to have a unique solution, the values of c can be -2/3 and -1/3.
Finally, to find the sum of all the real values of c that satisfy the conditions, we add up the values:
2 + 8 + (-2) + (-2/3) + (-1/3) = 15 - 5/3 = 40/3
So, the value of a+b is 40+3 = 43.