Determine all real numbers a such that the inequality |x2+2ax+3a|≤2 has exactly one solution in x.
Complete the square
f(x)=(x2+2ax+a2)+(3a−a2)=(x+a)2+(3a−a2)
So the minimum value is 3a-a^2
so 3a−a2=2
a=1,2
+21 
