If x 2 + 5 x + 5 x + 5 = 1 then the number of integers satisfying this equation is

Step-by-step solution

Given that, x 2 + 5 x + 5 x + 5 = 1 This is possible when x 2 + 5 x + 5 ≠ 0   &   x + 5 = 0 or x 2 + 5 x + 5 = 1 or x 2 + 5 x + 5 = - 1   &   x + 5 = even   integer Case1, x + 5 = 0 ⇒ x = - 5 And for this value of x , x 2 + 5 x + 5 ≠ 0 Case2, x 2 + 5 x + 5 = 1 ⇒ x 2 + 5 x + 4 = 0 ⇒ x + 1 x + 4 = 0 ⇒ x = - 1 ,   - 4 Case3, x 2 + 5 x + 5 = - 1 ⇒ x 2 + 5 x + 6 = 0 ⇒ x + 2 x + 3 = 0 ⇒ x = - 2 ,   - 3 But for x = - 2 , we ge