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.
You sure these are < and >, not <= and >=?
Solutions to such are open intervals.
Intersections of open intervals cannot be a single value for x.
ohw... sorry for that... for clarification, it's:
{(c-1)x^2+2cx+c+4<or=0
{ cx^2 + 2(c+1)x+(c+1)>or=0
5/4. Giving 9
no, its incorrect
19 is the correct answer
To find the sum of all real values of c for which the system of inequalities has a unique solution, we need to analyze each inequality separately and find the conditions under which they have a unique solution.
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 analyze the discriminant (the expression under the square root) of the quadratic equation:
Δ = b^2 - 4ac
Comparing this equation to our inequality, we have:
a = (c-1)
b = 2c
c = (c + 4)
Substituting these values into the discriminant equation, we get:
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 Δ must be equal to zero. Therefore:
-12c + 16 = 0
-12c = -16
c = -16 / -12
c = 4 / 3
So, the first inequality has a unique solution for c = 4 / 3.
Now, let's analyze the second inequality:
cx^2 + 2(c+1)x + (c+1) > 0
We can rewrite this inequality as:
cx^2 + 2cx + 2x + c + 1 > 0
Rearranging the terms, we have:
cx^2 + (2c+2)x + (c+1) > 0
Similarly, we can find the discriminant for this inequality:
Δ' = (2c+2)^2 - 4c(c+1)
= 4c^2 + 8c + 4 - 4c^2 - 4c
= 4c + 4
For a quadratic equation to have a unique solution, the discriminant Δ' must be equal to zero. Therefore:
4c + 4 = 0
4c = -4
c = -4 / 4
c = -1
The second inequality has a unique solution for c = -1.
Now, let's find the sum of all real values of c that satisfy both inequalities:
c = 4 / 3 and c = -1
The sum of these values is:
4 / 3 + (-1) = 4 / 3 - 3 / 3 = 1 / 3
Therefore, the sum of all real values of c is 1/3, which can be written as ab, where a = 1 and b = 3. So, the value of a + b is 1 + 3 = 4.
Hence, the value of a + b is 4.