To solve the quadratic equation \(14x^2 + 2x + 4 = 0\) using the perfect square trinomial pattern, we first need to arrange the equation into a form that can be factored as a perfect square.
Step 1: Divide the entire equation by 14 to simplify it:
\[ x^2 + \frac{1}{7}x + \frac{2}{7} = 0 \]
Step 2: Rewrite the equation:
\[ x^2 + \frac{1}{7}x = -\frac{2}{7} \]
Step 3: To complete the square, we need to add \(\left(\frac{b}{2}\right)^2\) to both sides, where \(b = \frac{1}{7}\):
\[ b = \frac{1}{7} \quad \Rightarrow \quad \frac{b}{2} = \frac{1}{14} \]
Now calculate \(\left(\frac{1}{14}\right)^2\):
\[ \left(\frac{1}{14}\right)^2 = \frac{1}{196} \]
Step 4: Add \(\frac{1}{196}\) to both sides:
\[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{2}{7} + \frac{1}{196} \]
To simplify the right side, we need a common denominator. The common denominator of 7 and 196 is 196. Therefore:
\[ -\frac{2}{7} = -\frac{2 \times 28}{7 \times 28} = -\frac{56}{196} \]
Adding these gives:
\[ -\frac{56}{196} + \frac{1}{196} = -\frac{55}{196} \]
Thus, the equation now looks like this:
\[ \left(x + \frac{1}{14}\right)^2 = -\frac{55}{196} \]
Step 5: Taking the square root of both sides gives:
\[ x + \frac{1}{14} = \pm \sqrt{-\frac{55}{196}} \]
The square root of a negative number indicates complex solutions. We can express it as:
\[ \sqrt{-\frac{55}{196}} = \frac{i\sqrt{55}}{14} \]
Step 6: Solve for \(x\):
\[ x + \frac{1}{14} = \pm \frac{i\sqrt{55}}{14} \]
Thus:
\[ x = -\frac{1}{14} \pm \frac{i\sqrt{55}}{14} \]
Step 7: Combine the terms:
\[ x = \frac{-1 \pm i\sqrt{55}}{14} \]
The solutions to the quadratic equation \(14x^2 + 2x + 4 = 0\) are:
\[ x = \frac{-1 + i\sqrt{55}}{14} \quad \text{and} \quad x = \frac{-1 - i\sqrt{55}}{14} \]