To solve the quadratic equation \( 14x^2 + 2x + 4 = 0 \) using the perfect square trinomial pattern, we can start by factoring out the greatest common factor (GCF) from the equation.
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Factor out the GCF (if applicable): The GCF of the coefficients \( 14 \), \( 2 \), and \( 4 \) is \( 2 \). So, we factor out \( 2 \):
\[ 2(7x^2 + x + 2) = 0 \]
Now we will solve the equation inside the parentheses:
\[ 7x^2 + x + 2 = 0 \]
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Rearrangement and completing the square: To complete the square for \( 7x^2 + x + 2 \), we can rearrange the quadratic part:
First, divide the entire equation by \( 7 \) (this will help with completing the square):
\[ x^2 + \frac{1}{7}x + \frac{2}{7} = 0 \]
Next, we will move \( \frac{2}{7} \) to the other side:
\[ x^2 + \frac{1}{7}x = -\frac{2}{7} \]
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Completing the square: To complete the square, take half of the coefficient of \( x \) (which is \( \frac{1}{7} \)), square it, and add it to both sides.
Half of \( \frac{1}{7} \) is \( \frac{1}{14} \). Squaring it gives:
\[ \left(\frac{1}{14}\right)^2 = \frac{1}{196} \]
Now add \( \frac{1}{196} \) to both sides:
\[ x^2 + \frac{1}{7}x + \frac{1}{196} = -\frac{2}{7} + \frac{1}{196} \]
To combine the fractions on the right side, we convert \( -\frac{2}{7} \) to have a denominator of \( 196 \):
\[ -\frac{2}{7} = -\frac{56}{196} \]
Now combine:
\[ -\frac{56}{196} + \frac{1}{196} = -\frac{55}{196} \]
So our equation now looks like this:
\[ \left(x + \frac{1}{14}\right)^2 = -\frac{55}{196} \]
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Taking the square root: Now we proceed by taking the square root of both sides:
\[ x + \frac{1}{14} = \pm \sqrt{-\frac{55}{196}} \]
The square root of a negative number involves imaginary numbers:
\[ x + \frac{1}{14} = \pm \frac{\sqrt{55}}{14}i \]
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Solve for \( x \): Finally, isolate \( x \):
\[ x = -\frac{1}{14} \pm \frac{\sqrt{55}}{14} i \]
Thus, the solutions to the equation \( 14x^2 + 2x + 4 = 0 \) are:
\[ x = -\frac{1}{14} + \frac{\sqrt{55}}{14} i \quad \text{and} \quad x = -\frac{1}{14} - \frac{\sqrt{55}}{14} i \]
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