If `alpha,beta` be the zeroes of the polynomial `2x^(2)+5x+k` such that `alpha^(2)+beta^(2)+alphabeta=(21)/(4)`, then k=?

Options

A. 3
B. `-3`
C. 2
D. `-2`

Answer

To find the value of \( k \) in the polynomial \( 2x^2 + 5x + k \) given that the zeros \( \alpha \) and \( \beta \) satisfy the equation \( \alpha^2 + \beta^2 + \alpha \beta = \frac{21}{4} \), we can follow these steps:

### Step 1: Identify the coefficients
The polynomial can be expressed in the standard form \( Ax^2 + Bx + C \), where:
- \( A = 2 \)
- \( B = 5 \)
- \( C = k \)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{B}{A} = -\frac{5}{2} \)
- The product of the roots \( \alpha \beta = \frac{C}{A} = \frac{k}{2} \)

### Step 3: Express \( \alpha^2 + \beta^2 \)
We can express \( \alpha^2 + \beta^2 \) using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting the values we have:
\[ \alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2\left(\frac{k}{2}\right) \]
Calculating \( \left(-\frac{5}{2}\right)^2 \):
\[ \alpha^2 + \beta^2 = \frac{25}{4} - k \]

### Step 4: Substitute into the given equation
Now, substitute \( \alpha^2 + \beta^2 \) into the equation \( \alpha^2 + \beta^2 + \alpha \beta = \frac{21}{4} \):
\[ \frac{25}{4} - k + \frac{k}{2} = \frac{21}{4} \]

### Step 5: Simplify the equation
To simplify, first combine the terms involving \( k \):
\[ \frac{25}{4} - k + \frac{k}{2} = \frac{21}{4} \]
Multiply through by 4 to eliminate the fractions:
\[ 25 - 4k + 2k = 21 \]
This simplifies to:
\[ 25 - 2k = 21 \]

### Step 6: Solve for \( k \)
Rearranging gives:
\[ 25 - 21 = 2k \]
\[ 4 = 2k \]
Dividing both sides by 2:
\[ k = 2 \]

### Final Answer
Thus, the value of \( k \) is \( \boxed{2} \).