Question

find three consecutive positive integers such that the product of the first and the third integer os 29 more than the second

Answer 1

5, 6, 7

Explanation:

let the 3 consecutive integers be n , n + 1 , n + 2 , then

n(n + 2) = n+ 1 + 29

n² + 2n = n + 30 ( subtract n + 30 from both sides )

n² + n - 30 = 0 ← in standard form

(n + 6)(n - 5) = 0 ← in factored form

equate each factor to zero and solve for n

n + 6 = 0 ⇒ n = - 6

n - 5 = 0 ⇒ n = 5

n > 0 , then n = 5

the 3 integers are 5, 5 + 1, 5 + 2 , that is

5, 6, 7