Consider the equation (x+p)(x+q)−k=0 x2+(p+q)x+pq−k=0 We get p+q=−(m+n) pq−k=mn...(i) Now, consider the equations whose roots are required to be found out. (x−m)(x−n)+k=0 x2−(m+n)x+mn+k=0 Substituting the value of ‘m’ and ‘n’ in the above equations, we get x2−(−(p+q))x+pq−k+k=0 x2+(p+q)x+pq=0 (x+p)(x+q)=0 x=−p,x=−q