To solve the quadratic equation \( \frac{1}{4}x^2 + 2x + 4 = 0 \) using the perfect square trinomial pattern, we need to manipulate the equation into a form that fits \( (ax + b)^2 = c \).
First, let's multiply the entire equation by 4 to eliminate the fraction:
\[ 4 \left( \frac{1}{4}x^2 + 2x + 4 \right) = 0 \]
This simplifies to:
\[ x^2 + 8x + 16 = 0 \]
Next, we can recognize that the left-hand side can be expressed as a perfect square:
\[ (x + 4)^2 = 0 \]
Now, to solve for \( x \), we can take the square root of both sides:
\[ x + 4 = 0 \]
Subtracting 4 from both sides gives:
\[ x = -4 \]
So, the solution to the equation is \( x = -4 \).
Why is -4 the correct answer?
The answer \( x = -4 \) is correct because when we substitute it back into the original equation, it satisfies the equality:
- Substitute \( x = -4 \) into the original equation: \[ \frac{1}{4}(-4)^2 + 2(-4) + 4 = 0 \]
- Calculate: \[ \frac{1}{4} \cdot 16 - 8 + 4 = 0 \] \[ 4 - 8 + 4 = 0 \] \[ 0 = 0 \]
Since the left side equals the right side, it confirms that \( x = -4 \) is indeed the correct solution to the quadratic equation.