Question

Solve the following quadratic equation using the perfect square trinomial pattern: 14x2+2x+4=0

Answer 1

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:

\[
x = \frac{-1 + i\sqrt{55}}{14} \quad \text{and} \quad x = \frac{-1 - i\sqrt{55}}{14}
\]