If α, β are the zeros of the polynomial f(x)

Given: $f(x)=x^{2}-5 x+k$

The co-efficients are $a=1, b=-5$ and $c=k$.

$\therefore \alpha+\beta=\frac{-b}{a}$

$=>\alpha+\beta=-\frac{(-5)}{1}$

$=>\alpha+\beta=5 \quad \ldots(1)$

Also, $\alpha-\beta=1 \quad \ldots(2)$

From (1) & (2), we get:

$2 \alpha=6$

$=>\alpha=3$

Pu tting the value of $\alpha$ in $(1)$, we get $\beta=2$.

Now, $\alpha \beta=\frac{c}{a}$

$=>3 \times 2=\frac{k}{1}$

$\therefore k=6$