To solve the quadratic equation \( 14x^2 + 2x + 4 = 0 \) using the perfect square trinomial pattern, we first rewrite the equation in standard form:
-
Divide all terms by 14 to simplify: \[ x^2 + \frac{2}{14}x + \frac{4}{14} = 0 \] which simplifies to: \[ x^2 + \frac{1}{7}x + \frac{2}{7} = 0 \]
-
Move the constant term to the other side: \[ x^2 + \frac{1}{7}x = -\frac{2}{7} \]
-
To complete the square, take half of the coefficient of \( x \), square it, and add it to both sides:
- The coefficient of \( x \) is \( \frac{1}{7} \). Half of it is \( \frac{1}{14} \), and squaring it gives \( \left(\frac{1}{14}\right)^2 = \frac{1}{196} \).
Now add \( \frac{1}{196} \) to both sides: \[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{2}{7} + \frac{1}{196} \]
-
Simplify the right side:
- To combine \( -\frac{2}{7} \) and \( \frac{1}{196} \), convert \( -\frac{2}{7} \) to a fraction with a denominator of 196: \[ -\frac{2}{7} = -\frac{2 \times 28}{7 \times 28} = -\frac{56}{196} \]
- Now combine: \[ -\frac{56}{196} + \frac{1}{196} = -\frac{55}{196} \]
-
Now we have: \[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{55}{196} \] which can be rewritten as: \[ \left(x + \frac{1}{14}\right)^2 = -\frac{55}{196} \]
-
Since the right side is negative, we take the square root of both sides: \[ x + \frac{1}{14} = \pm \sqrt{-\frac{55}{196}} = \pm \frac{\sqrt{55}i}{14} \]
-
Now solve for \( x \): \[ x = -\frac{1}{14} \pm \frac{\sqrt{55}i}{14} \]
Thus, the solutions to the equation \( 14x^2 + 2x + 4 = 0 \) are: \[ x = -\frac{1}{14} + \frac{\sqrt{55}i}{14}, \quad x = -\frac{1}{14} - \frac{\sqrt{55}i}{14} \]
1 answer