To solve the quadratic equation \( \frac{1}{4}x^2 + 2x + 4 = 0 \) using the perfect square trinomial pattern, we first rearrange the equation.
Let's start by moving the constant to the other side:
\[ \frac{1}{4}x^2 + 2x = -4 \]
Next, we want to express the left-hand side as a perfect square trinomial. First, we can factor out \( \frac{1}{4} \) from the left side:
\[ \frac{1}{4}(x^2 + 8x) = -4 \]
Now, we will complete the square for the expression \( x^2 + 8x \). To complete the square, we take half of the coefficient of \( x \) (which is 8), square it, and add it inside the parentheses. Half of 8 is 4, and \( 4^2 = 16 \).
We rewrite the expression:
\[ x^2 + 8x + 16 - 16 \]
Thus,
\[ \frac{1}{4}((x + 4)^2 - 16) = -4 \]
Now simplify the left-hand side:
\[ \frac{1}{4}(x + 4)^2 - 4 = -4 \]
To isolate the square, add 4 to both sides:
\[ \frac{1}{4}(x + 4)^2 = 0 \]
Now, multiply both sides by 4:
\[ (x + 4)^2 = 0 \]
Taking the square root of both sides gives:
\[ x + 4 = 0 \]
Solving for \( x \):
\[ x = -4 \]
So the solution to the quadratic equation \( \frac{1}{4}x^2 + 2x + 4 = 0 \) is:
\[ \boxed{-4} \]