using the relationship between the zeroes of the quadratic polynomial.

Sum of zeroes $=\frac{-(\text { coef ficient of } x)}{\text { coefficient of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coef ficient of } x^{2}}$

$\therefore \alpha+\beta=\frac{-(-5)}{1}$ and $\alpha \beta=\frac{k}{1}$

$\Rightarrow \alpha+\beta=5$ and $\alpha \beta=\frac{k}{1}$

Solving $\alpha-\beta=1$ and $\alpha+\beta=5$

$\alpha=3$ and $\beta=2$

Substituting these values in $\alpha \beta=\frac{k}{1}$,

$k=6$