If c, d are the roots of the equation (x – a) (x – b) – k = 0, then the roots of the equation (x – c) (x – d) + k = 0 are

we have,

$\Rightarrow {\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}-\mathrm{k}=0$

$\Rightarrow {\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}-\mathrm{k}=0----\left(1\right)$

Since the roots of Eq. (1) are c and d

$\begin{array}{rl}\therefore \mathrm{c}+\mathrm{d}& =\mathrm{a}+\mathrm{b},----\left(2\right)\\ \mathrm{and} \mathrm{cd}& =\mathrm{ab}-\mathrm{k}---\left(3\right)\\ \mathrm{Now}, (\mathrm{x}-\mathrm{c})(\mathrm{x}-\mathrm{d})+\mathrm{k}& =0\\ \Rightarrow {\mathrm{x}}^2-(\mathrm{c}+\mathrm{d})\mathrm{x}+\mathrm{cd}+\mathrm{k}& =0\\ \Rightarrow {\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}& =0\end{array}$

[Putting the values of a + b and ab from Eqs (2) and (3)] 

$\Rightarrow (\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})=0\Rightarrow \mathrm{x}=\mathrm{a},\mathrm{b}$