(x2−11x+29)(x2−8x+7)=1
Let a=x2−11x+29 and b=x2−8x+7
Case 1: ab=1 when a=1
x2−11x+29=1
⇒x2−11x+28=0⇒x=4,7
Case 2: ab=1 when b=0
x2−8x+7=0⇒x=1,7
Case 3: ab=1 when a=−1 and b is an even number.
x2−11x+29=−1
⇒x2−11x+30=0⇒x=5,6
For x=5, LHS =(−1)−8=1=RHS
For x=6, LHS =(−1)−5=−1≠RHS
Therefore, distinct roots are 1,4,5,7
Sum of distinct roots =4+7+1+5=17