find the values `k` for which the quadratic equation `2x^2 + Kx +3=0` has two real equal roots

Options

A. lies between 1 and 2
B. lies between 2 and 3
C. lies between -1 and 0
D. does not exist

Explanation

To find the values of \( k \) for which the quadratic equation \( 2x^2 + kx + 3 = 0 \) has two real equal roots, we need to use the condition related to the discriminant of a quadratic equation.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \).
Here, \( a = 2 \), \( b = k \), and \( c = 3 \).

2. **Write the formula for the discriminant**:
The discriminant \( D \) of a quadratic equation is given by:
\[ D = b^2 - 4ac \]

3. **Substitute the coefficients into the discriminant formula**:
Substitute \( a \), \( b \), and \( c \) into the discriminant formula:
\[ D = k^2 - 4 \cdot 2 \cdot 3 \]

4. **Simplify the expression**:
Calculate \( 4 \cdot 2 \cdot 3 \):
\[ D = k^2 - 24 \]

5. **Set the discriminant equal to zero for equal roots**:
For the quadratic equation to have two real equal roots, the discriminant must be equal to zero:
\[ k^2 - 24 = 0 \]

6. **Solve for \( k \)**:
Rearranging the equation gives:
\[ k^2 = 24 \]
Taking the square root of both sides, we find:
\[ k = \pm \sqrt{24} \]
Simplifying \( \sqrt{24} \):
\[ k = \pm 2\sqrt{6} \]

### Final Answer:
The values of \( k \) for which the quadratic equation \( 2x^2 + kx + 3 = 0 \) has two real equal roots are:
\[ k = 2\sqrt{6} \quad \text{and} \quad k = -2\sqrt{6} \]