$$ 4^{x^2} - 5 \cdot 2^{x^2} + 4 = 0 $$ — \begin{align*} (2^{2})^{x^2} - 5 \cdot 2^{x^2} + 4 & = 0 \\ (2^{x^2})^{2} - 5 \cdot 2^{x^2} + 4 & = 0 \end{align*} Seja $ 2^{x^2} = a , \: a >0$ \begin{align*} a^2 - 5 a + 4 & = 0 \\ (a-4)(a-1) & = 0 \end{align*} $a = 1$: $$ 2^{x^2} = 1 \Leftrightarrow x = 0 $$ $ a = 4 $: $$ 2^{x^2} = 4 \Leftrightarrow x^2 = 2 \Leftrightarrow x = \pm \sqrt{2} $$ A única raiz positiva é $ \sqrt{2}$ $$ S = \sqrt{2} $$ Alternativa correta: $\boxed{\mathrm{C}}$ $$ \boxed{ S = \sqrt{2}} $$