\(9^x - (4a)3^x + 4 - a^2 = 0\)
\(D = 0\)
\(3^x = y\)
\(y^2 - 4ay + (4 - a^2) = 0\)
\(\Rightarrow b^2 - 4ac = D\)
\(\Rightarrow 16a^2 - 4(4 - a^2) = 0\)
\(\Rightarrow 16a^2 - 16+ 4a^2 = 0\)
\(\Rightarrow 20a^2 = 16\)
\(\Rightarrow a = \pm \frac 2{\sqrt 5}\)
\(\therefore 3^x = \frac{-(-4a)}{2(1)} = \frac{4a}2 = 2a\)
⇒ \(x = \log_3 (2a)\)
\(\therefore a = \frac 2{\sqrt 5}\)