Determine all real numbers a such that the inequality \(|x^2 + 2ax + 3a|\le2\) has exactly one solution in x.
Complete the square
\(f(x) = (x^2+2ax+a^2) + (3a-a^2) = (x+a)^2 + (3a-a^2)\)
So the minimum value is 3a-a^2
so \(3a - a^2 = 2\)
a=1,2
+21 
