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

Options

A. `(-3sqrt3, 4sqrt3)`
B. `[-4sqrt3, 4sqrt3]`
C. `((-1)/(4), (1)/(4))`
D. `((1)/(4), (-1)/(4))`

Answer

To solve the inequality \(4x^2 + kx + 3 \geq 0\) for all real values of \(x\), we need to analyze the quadratic expression. The key is to ensure that the discriminant of the quadratic is non-positive, which guarantees that the quadratic does not cross the x-axis and remains non-negative.

### Step-by-step solution:

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

2. **Calculate the discriminant**:
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = k^2 - 4 \cdot 4 \cdot 3 = k^2 - 48 \]

3. **Set the discriminant condition**:
For the quadratic to be non-negative for all \(x\), the discriminant must be less than or equal to zero:
\[ k^2 - 48 \leq 0 \]

4. **Solve the inequality**:
Rearranging the inequality gives:
\[ k^2 \leq 48 \]
Taking the square root of both sides, we find:
\[ -\sqrt{48} \leq k \leq \sqrt{48} \]
Since \(\sqrt{48} = 4\sqrt{3}\), we can express this as:
\[ -4\sqrt{3} \leq k \leq 4\sqrt{3} \]

5. **Conclusion**:
Therefore, the values of \(k\) lie in the interval:
\[ k \in [-4\sqrt{3}, 4\sqrt{3}] \]

### Final Answer:
The interval for \(k\) is \([-4\sqrt{3}, 4\sqrt{3}]\).