word problem

Question

Find the three consecutive integers such that the positive difference of the square of the first two is 8 more than the  third.              

Answer 1

Let the integers be n-1, n and n+1.

n²-(n-1)²=n+1+8 or (n+1)²+8. The question isn't clear which of these applies so I'll give separate answers for each.

(a) n²-(n-1)²=n+1+8,

(n-n+1)(n+n-1)=n+9,

2n-1=n+9, n=10, so the integers are 9, 10, 11.

(b) n²-(n-1)²=(n+1)²+8,

2n-1=n²+2n+9, -1=n²+9, n²=-10 clearly has no real solution, so (a) applies.

SOLUTION 9, 10, 11.