Problema Solution
Find 3 consecutive integers such that the product of the largest two is 20 more than the square of the smallest integer.
Answer provided by our tutors
let the three consecutive integers be
x - 1, x and x + 1,
where x is integer
also x - 1 < x < x + 1
the product of the largest two is 20 more than the square of the smallest integer
x(x + 1) = 20 + (x - 1)^2
by solving we find
x = 7
click here to see the step by step solution of the equations
x - 1 = 7 - 1 = 6
x + 1 = 7 + 1 = 8
indeed 7*8 = 20 + 6^2
the numbers are 6,7 and 8.