Question

three consecutive integersare such that square of first number and product of remaining two add up to 121. find the numbers.

Answer 1

7,8 And 9 I am pretty sure about it.

Answer 2

Let the three consecutive no`s be x, x+1, x+2
Given that x² + [(x+1)(x+2)] = 121
⇒x² + [x² + 2x + x +2] = 121
⇒2x² + 3x +2 - 121 = 0
⇒2x² + 3x -119 = 0
⇒2x² +17x -14x - 119 = 0
⇒x(2x + 17) - 7(2x + 17)
⇒ (2x +17) (x-7) = 0
⇒2x + 17 = 0       or       x-7 = 0
⇒2x = - 17           or       x = 7
⇒x   = -17/2         or       x =7
So the three consecutive no`s are 7,8,9.