If 3 is a root of `x^(2)+kx-24=0` it is also a root of

Options

A. `x^(2)+5x+k=0`
B. `x^(2)+kx+24=0`
C. `x^(2)-kx+6=0`
D. `x^(2)-5x+k=0`

Answer

To solve the problem, we need to find the value of \( k \) given that 3 is a root of the quadratic equation \( x^2 + kx - 24 = 0 \). Then, we will check which of the provided equations also has 3 as a root.

### Step-by-Step Solution:

1. **Substituting the root into the equation**:
Since 3 is a root of the equation \( x^2 + kx - 24 = 0 \), we can substitute \( x = 3 \) into the equation:
\[ 3^2 + k(3) - 24 = 0 \]

2. **Calculating \( 3^2 \)**:
Calculate \( 3^2 \):
\[ 9 + 3k - 24 = 0 \]

3. **Simplifying the equation**:
Rearranging the equation gives:
\[ 3k - 15 = 0 \]

4. **Solving for \( k \)**:
Now, solve for \( k \):
\[ 3k = 15 \implies k = \frac{15}{3} = 5 \]

5. **Substituting \( k \) back into the equations**:
Now that we have \( k = 5 \), we can substitute it into the given equations to check which one also has 3 as a root.

6. **Checking each equation**:
- **Equation 1**: \( x^2 + 5x + k = 0 \) becomes \( x^2 + 5x + 5 = 0 \)
- Substitute \( x = 3 \):
\[ 3^2 + 5(3) + 5 = 9 + 15 + 5 = 29 \quad (\text{not a root}) \]

- **Equation 2**: \( x^2 + kx + 24 = 0 \) becomes \( x^2 + 5x + 24 = 0 \)
- Substitute \( x = 3 \):
\[ 3^2 + 5(3) + 24 = 9 + 15 + 24 = 48 \quad (\text{not a root}) \]

- **Equation 3**: \( x^2 - kx + 6 = 0 \) becomes \( x^2 - 5x + 6 = 0 \)
- Substitute \( x = 3 \):
\[ 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0 \quad (\text{is a root}) \]

- **Equation 4**: \( x^2 - 5x + k = 0 \) becomes \( x^2 - 5x + 5 = 0 \)
- Substitute \( x = 3 \):
\[ 3^2 - 5(3) + 5 = 9 - 15 + 5 = -1 \quad (\text{not a root}) \]

7. **Conclusion**:
The equation that has 3 as a root is:
\[ x^2 - 5x + 6 = 0 \]

### Final Answer:
The equation that also has 3 as a root is \( x^2 - 5x + 6 = 0 \).