If `alpha` and `beta` are zeros of `y^(2)+5y+m`, find the value of `m` such that `(alpha+beta)^(2)-alphabeta=24`

Answer

To solve the problem, we need to find the value of \( m \) such that the condition \( (\alpha + \beta)^2 - \alpha \beta = 24 \) holds true, where \( \alpha \) and \( \beta \) are the zeros of the polynomial \( y^2 + 5y + m \).

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial**:
The given polynomial is \( y^2 + 5y + m \). Here, we can identify:
- \( a = 1 \)
- \( b = 5 \)
- \( c = m \)

2. **Use Vieta's formulas**:
According to Vieta's formulas, for a quadratic polynomial \( ay^2 + by + c \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Applying this to our polynomial:
- \( \alpha + \beta = -\frac{5}{1} = -5 \)
- \( \alpha \beta = \frac{m}{1} = m \)

3. **Substitute into the given condition**:
We are given the condition:
\[ (\alpha + \beta)^2 - \alpha \beta = 24 \]
Substituting the values we found:
\[ (-5)^2 - m = 24 \]

4. **Calculate \( (-5)^2 \)**:
\[ 25 - m = 24 \]

5. **Solve for \( m \)**:
Rearranging the equation gives:
\[ 25 - 24 = m \]
\[ m = 1 \]

### Final Answer:
The value of \( m \) is \( 1 \).