Consider the following system of inequalities:

{(c−1)x^2+2cx+c+4≤0
cx^2+2(c+1)x+(c+1)≥0
The sum of all real values of c, such that the system has a unique solution, can be written as ab, 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 solve this problem, we need to analyze each inequality separately and find the values of "c" for which each inequality has a unique solution. Then, we can find the common values of "c" that satisfy both inequalities.

Let's start by analyzing the first inequality:

(c-1)x^2 + 2cx + c + 4 ≤ 0

To find the values of "c" for which this inequality has a unique solution, we need to find the discriminant of the quadratic equation. The discriminant, denoted by Δ, is given by Δ = b^2 - 4ac.

In this case, a = c - 1, b = 2c, and c = c + 4. Substituting these values into the discriminant formula, we have:

Δ = (2c)^2 - 4(c - 1)(c + 4)
= 4c^2 - 4(c^2 + 3c - 4)
= 4c^2 - 4c^2 - 12c + 16
= -12c + 16

For a quadratic equation to have a unique solution, the discriminant must be equal to zero. Therefore, we can set Δ = 0 and solve for "c":

-12c + 16 = 0
-12c = -16
c = 4/3

So, the first inequality has a unique solution when c = 4/3.

Now, let's analyze the second inequality:

cx^2 + 2(c+1)x + (c+1) ≥ 0

For this inequality to have a unique solution, the quadratic equation must either be always positive or always negative. To determine when this happens, we can examine the leading coefficient "c" and the discriminant.

Since "c" can be negative according to the given details and assumptions, we have two cases to consider:

Case 1: c > 0
In this case, the leading coefficient "c" is positive. If the discriminant is less than zero (Δ < 0), the quadratic equation will always be positive, satisfying the inequality. Therefore, Δ < 0.

Δ = (2(c+1))^2 - 4c(c+1)
= 4(c^2 + 2c + 1) - 4c(c+1)
= 4c^2 + 8c + 4 - 4c^2 - 4c
= 4c + 4

Setting Δ < 0, we have:

4c + 4 < 0
4c < -4
c < -1

In this case, the inequality has a unique solution for c < -1.

Case 2: c < 0
In this case, the leading coefficient "c" is negative. If the discriminant is greater than or equal to zero (Δ ≥ 0), the quadratic equation will always be negative, satisfying the inequality. Therefore, Δ ≥ 0.

Δ = (2(c+1))^2 - 4c(c+1)
= 4(c^2 + 2c + 1) - 4c(c+1)
= 4c^2 + 8c + 4 - 4c^2 - 4c
= 4c + 4

Setting Δ ≥ 0, we have:

4c + 4 ≥ 0
4c ≥ -4
c ≥ -1

In this case, the inequality has a unique solution for c ≥ -1.

Therefore, the second inequality has a unique solution when c < -1 or c ≥ -1.

To find the values of "c" that satisfy both inequalities, we need to find the intersection of the solutions obtained from each inequality:

For the first inequality: c = 4/3
For the second inequality: c < -1 or c ≥ -1

The only value of "c" that satisfies both inequalities is c = 4/3.

Finally, we can write the sum of all real values of "c" as ab, where a = 4 and b = 3. Therefore, a + b = 4 + 3 = 7.

So, the value of a + b is 7.