If `3x^(2)+4kx+1 gt 0` for all real values of x, then k lies in the interval

Options

A. `((-sqrt(3))/(2), (sqrt(3))/(2))`
B. `((-1)/(4), (1)/(4))`
C. `[(-sqrt(3))/(2), (sqrt(3))/(2)]`
D. `((-1)/(2), (1)/(2))`

Answer

To solve the inequality \(3x^2 + 4kx + 1 > 0\) for all real values of \(x\), we need to ensure that the quadratic expression does not have any real roots. This can be achieved by ensuring that the discriminant of the quadratic is less than zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The quadratic can be compared to the standard form \(ax^2 + bx + c\):
- \(a = 3\)
- \(b = 4k\)
- \(c = 1\)

2. **Calculate the discriminant**:
The discriminant \(D\) of a quadratic equation is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (4k)^2 - 4(3)(1) = 16k^2 - 12 \]

3. **Set the discriminant less than zero**:
For the quadratic to be positive for all \(x\), we need:
\[ 16k^2 - 12 < 0 \]

4. **Solve the inequality**:
Rearranging gives:
\[ 16k^2 < 12 \]
Dividing both sides by 16:
\[ k^2 < \frac{12}{16} = \frac{3}{4} \]

5. **Take the square root**:
Taking the square root of both sides (and remembering to consider both the positive and negative roots):
\[ -\sqrt{\frac{3}{4}} < k < \sqrt{\frac{3}{4}} \]
Simplifying gives:
\[ -\frac{\sqrt{3}}{2} < k < \frac{\sqrt{3}}{2} \]

6. **Final interval**:
Thus, the values of \(k\) lie in the interval:
\[ k \in \left(-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right) \]

### Conclusion:
The final answer is that \(k\) lies in the interval \(\left(-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right)\).