SOLUTION: Find three consecutive integers such that the square of the smallest is 29 less than the product of the larger two.

SOLUTION: Find three consecutive integers such that the square of the smallest is 29 less than the product of the larger two.

Algebra.Com

Question 1131637: Find three consecutive integers such that the square of the smallest is 29 less than the product of the larger two.
Answer by ikleyn(52834) (Show Source): You can put this solution on YOUR website!
.

n^2 = (n+1)*(n+2) - 29


n^2 = n^2 + n + 2n + 2 - 29


3n = 27  ====>  n = 9.


Answer.  9, 10 and 11.

Solved.

RELATED QUESTIONS
Find three consecutive even integers such that three times the sum of all three is... (answered by checkley79)

Find three consecutive integers such that the product of the two largest is 20 more than... (answered by CubeyThePenguin)

The smallest of three consecutive integers is 18 less than the sum of the two larger... (answered by stanbon)

Find four consecutive integers such that the product of the two larger integers is 22... (answered by jorel1380)

Find three consecutive integers such that the product of the two largest is 20 more than... (answered by jorel1380)

find three consecutive odd integers such that the sum of all three is 42 less than the... (answered by KMST)

Find three consecutive positive integers such that twice the product of the two smaller... (answered by JBarnum)

Find three consecutive natural numbers such that the square of the smallest number is 65... (answered by Fombitz)

Find two consecutive integers whose product is 6 less than the square of the larger... (answered by drk)