To solve the quadratic equation \(14x^2 + 2x + 4 = 0\) using the perfect square trinomial pattern, we first want to rewrite the equation in a more manageable form.
First, we can divide the entire equation by 2 for simplicity:
\[ 7x^2 + x + 2 = 0 \]
Next, we can rearrange this in the form of \(ax^2 + bx + c\) and focus on completing the square for the quadratic expression.
We can express it as:
\[ 7x^2 + x = -2 \]
Now, factor out the coefficient of \(x^2\) (which is 7) from the left side:
\[ 7\left(x^2 + \frac{1}{7}x\right) = -2 \]
Next, we need to complete the square inside the parentheses. The term we need to complete the square is found from \(\left(\frac{b}{2}\right)^2\), where \(b\) is the coefficient of \(x\):
\[ \left(\frac{1}{14}\right)^2 = \frac{1}{196} \]
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 rearranged to become:
\[ 7\left(\left(x + \frac{1}{14}\right)^2 - \frac{1}{196}\right) = -2 \]
Distributing the 7 gives:
\[ 7\left(x + \frac{1}{14}\right)^2 - \frac{7}{196} = -2 \]
Simplifying \(\frac{7}{196}\):
\[ 7\left(x + \frac{1}{14}\right)^2 - \frac{1}{28} = -2 \]
Adding \(\frac{1}{28}\) to both sides gives:
\[ 7\left(x + \frac{1}{14}\right)^2 = -2 + \frac{1}{28} \]
To combine the right side, convert -2 to have a common denominator of 28:
\[ -2 = -\frac{56}{28} \]
Thus, the expression becomes:
\[ 7\left(x + \frac{1}{14}\right)^2 = -\frac{56}{28} + \frac{1}{28} = -\frac{55}{28} \]
Dividing both sides by 7:
\[ \left(x + \frac{1}{14}\right)^2 = -\frac{55}{196} \]
At this point, we see that we have a negative value under the square root, indicating that there are no real solutions. Therefore, we can conclude that the quadratic equation has complex solutions.
Taking the square root of both sides gives:
\[ x + \frac{1}{14} = \pm i\sqrt{\frac{55}{196}} \]
Which simplifies to:
\[ x + \frac{1}{14} = \pm \frac{i\sqrt{55}}{14} \]
Subtracting \(\frac{1}{14}\) from both sides yields:
\[ x = -\frac{1}{14} \pm \frac{i\sqrt{55}}{14} \]
So the two complex solutions are:
\[ x = -\frac{1}{14} + \frac{i\sqrt{55}}{14}, \quad \text{and} \quad x = -\frac{1}{14} - \frac{i\sqrt{55}}{14} \]
Therefore, the solution in terms of real and imaginary parts is:
\[ \boxed{-\frac{1}{14} \pm \frac{i\sqrt{55}}{14}} \]
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