To solve the equation \(4x^2 + 3x + 1 = 0\), we will use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
Here, \(a = 4\), \(b = 3\), and \(c = 1\).
First, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = 3^2 = 9 \] \[ 4ac = 4 \cdot 4 \cdot 1 = 16 \] \[ b^2 - 4ac = 9 - 16 = -7 \]
Since the discriminant is negative, the solutions will be complex numbers.
Now we substitute \(a\), \(b\), and the discriminant back into the quadratic formula:
\[ x = \frac{{-3 \pm \sqrt{{-7}}}}{2 \cdot 4} \]
The square root of \(-7\) can be expressed as \(i\sqrt{7}\):
\[ x = \frac{{-3 \pm i\sqrt{7}}}{8} \]
Now we can express the two complex roots:
- \(x_1 = \frac{{-3 + i\sqrt{7}}}{8}\)
- \(x_2 = \frac{{-3 - i\sqrt{7}}}{8}\)
Thus, the solutions to the equation \(4x^2 + 3x + 1 = 0\) are:
A. \(\frac{{-3 + i\sqrt{7}}}{8}, \frac{{-3 - i\sqrt{7}}}{8}\)
So, the correct answer is A.
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