To solve the quadratic equation \( 14x^2 + 2x + 4 = 0 \) using the perfect square trinomial pattern, we need to write it in the form \( (ax + b)^2 = c \).
First, we can simplify the equation by dividing all terms by 2:
\[ 7x^2 + x + 2 = 0 \]
Next, we need to isolate the \( x^2 \) and \( x \) terms. We can rewrite the equation as:
\[ 7x^2 + x = -2 \]
We can factor out the 7 from the \( x^2 \) term:
\[ 7 \left( x^2 + \frac{1}{7}x \right) = -2 \]
Now, in order to complete the square, we need to find the term that completes the square. The coefficient of \( x \) is \( \frac{1}{7} \). To complete the square, we need to take half of this coefficient, square it, and add it inside the parentheses:
- Half of \( \frac{1}{7} \) is \( \frac{1}{14} \).
- Squaring \( \frac{1}{14} \) gives \( \left( \frac{1}{14} \right)^2 = \frac{1}{196} \).
Now we add and subtract \( \frac{1}{196} \) inside the parentheses:
\[ 7 \left( x^2 + \frac{1}{7}x + \frac{1}{196} - \frac{1}{196} \right) = -2 \]
This can be rewritten as:
\[ 7 \left( \left( x + \frac{1}{14} \right)^2 - \frac{1}{196} \right) = -2 \]
Next, distribute the 7:
\[ 7 \left( x + \frac{1}{14} \right)^2 - \frac{7}{196} = -2 \]
Since \( -\frac{7}{196} = -\frac{1}{28} \):
\[ 7 \left( x + \frac{1}{14} \right)^2 - \frac{1}{28} = -2 \]
Now add \( \frac{1}{28} \) to both sides:
\[ 7 \left( x + \frac{1}{14} \right)^2 = -2 + \frac{1}{28} \]
To combine the fraction on the right-hand side, convert -2 into a fraction:
\[ -2 = -\frac{56}{28} \]
Now we have:
\[ 7 \left( x + \frac{1}{14} \right)^2 = -\frac{56}{28} + \frac{1}{28} = -\frac{55}{28} \]
Next, we divide by 7:
\[ \left( x + \frac{1}{14} \right)^2 = -\frac{55}{28 \times 7} = -\frac{55}{196} \]
This equation tells us that a square equals a negative number, indicating no real solutions exist. Instead, we can break down to find complex solutions.
Taking the square root of both sides:
\[ x + \frac{1}{14} = \pm i \sqrt{\frac{55}{196}} \]
This simplifies to:
\[ x + \frac{1}{14} = \pm i \frac{\sqrt{55}}{14} \]
Now, solving for \( x \):
\[ x = -\frac{1}{14} \pm i \frac{\sqrt{55}}{14} \]
Thus, the solutions to the quadratic equation \( 14x^2 + 2x + 4 = 0 \) are:
\[ x = -\frac{1}{14} + i \frac{\sqrt{55}}{14}, \quad x = -\frac{1}{14} - i \frac{\sqrt{55}}{14} \]