If the sum of squares of zeros of the polynomial `x^(2)-8x+k` is `40`, find the value of `k`.

Answer

To find the value of \( k \) in the polynomial \( x^2 - 8x + k \) given that the sum of squares of its zeros is 40, follow these steps:

1. **Identify the polynomial and its coefficients**:
The polynomial is \( x^2 - 8x + k \).
Here, \( a = 1 \), \( b = -8 \), and \( c = k \).

2. **Recall the relationships for the sum and product of the roots**:
For a quadratic polynomial \( ax^2 + bx + c \):
- Sum of the roots (zeros), \( \alpha + \beta = -\frac{b}{a} \)
- Product of the roots (zeros), \( \alpha \beta = \frac{c}{a} \)

3. **Calculate the sum and product of the roots**:
- Sum of the roots: \( \alpha + \beta = -\frac{-8}{1} = 8 \)
- Product of the roots: \( \alpha \beta = \frac{k}{1} = k \)

4. **Use the given condition about the sum of squares of the roots**:
The sum of squares of the roots is given as 40. We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the known values:
\[ \alpha^2 + \beta^2 = 8^2 - 2k \]
\[ \alpha^2 + \beta^2 = 64 - 2k \]

5. **Set up the equation with the given sum of squares**:
According to the problem, \( \alpha^2 + \beta^2 = 40 \). So:
\[ 64 - 2k = 40 \]

6. **Solve for \( k \)**:
\[ 64 - 40 = 2k \]
\[ 24 = 2k \]
\[ k = \frac{24}{2} = 12 \]

Therefore, the value of \( k \) is \( 12 \).