How do you find two consecutive odd integers such that the difference of their squares is 160?

1 Answer
Feb 1, 2016

Answer:

Set up an equation using #2n+1# as the general form of an odd number to find the integers to be #39# and #41#

Explanation:

Any odd integer may be represented as #2n+1# for some integer #n#. Let's let #n# be such that #2n+1# is the lesser of the two consecutive odd integers. Then we have the equation

#(2n+3)^2-(2n+1)^2 = 160#

#=>4n^2+12n+9-4n^2-4n-1 = 160#

#=>8n = 152#

#=> n = 19#

#:. 2n+1 = 39# and #2n+3 = 41#

Thus the two consecutive odd integers with the desired property are #39# and #41#.