Find the value of `k`, so that the equation `2x^2+kx-5=0` and `x^2-3x-4=0` may have one root in common.

Options

A. `-3`
B. `-1`
C. `-(27)/(4)`
D. `(27)/(4)`

Explanation

To find the value of \( k \) such that the equations \( 2x^2 + kx - 5 = 0 \) and \( x^2 - 3x - 4 = 0 \) have one root in common, we will follow these steps:

### Step 1: Solve the second equation for its roots
We start with the equation:
\[ x^2 - 3x - 4 = 0 \]
We can factor this equation. We want two numbers that multiply to \(-4\) (the constant term) and add to \(-3\) (the coefficient of \(x\)). The numbers \(-4\) and \(1\) satisfy these conditions.

Thus, we can factor the equation as:
\[ (x - 4)(x + 1) = 0 \]
Setting each factor to zero gives us the roots:
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
So, the roots of the second equation are \( x = 4 \) and \( x = -1 \).

### Step 2: Substitute the common root into the first equation
Now we will substitute each of these roots into the first equation \( 2x^2 + kx - 5 = 0 \) to find the corresponding values of \( k \).

#### Case 1: Using \( x = 4 \)
Substituting \( x = 4 \):
\[ 2(4^2) + k(4) - 5 = 0 \]
Calculating \( 4^2 \):
\[ 2(16) + 4k - 5 = 0 \]
This simplifies to:
\[ 32 + 4k - 5 = 0 \]
\[ 4k + 27 = 0 \]
Solving for \( k \):
\[ 4k = -27 \quad \Rightarrow \quad k = -\frac{27}{4} \]

#### Case 2: Using \( x = -1 \)
Now substituting \( x = -1 \):
\[ 2(-1)^2 + k(-1) - 5 = 0 \]
Calculating \( (-1)^2 \):
\[ 2(1) - k - 5 = 0 \]
This simplifies to:
\[ 2 - k - 5 = 0 \]
\[ -k - 3 = 0 \]
Solving for \( k \):
\[ -k = 3 \quad \Rightarrow \quad k = -3 \]

### Conclusion
The values of \( k \) that allow the two equations to have one root in common are:
\[ k = -\frac{27}{4} \quad \text{and} \quad k = -3 \]