You sure these are < and >, not <= and >=?
Solutions to such are open intervals.
Intersections of open intervals cannot be a single value for x.
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.
5 answers
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
{(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