The sum of the square of two consecutive positive odd integers is 202, how do you find the integers?
9 , 11
let n be a positive odd integer
then the next consecutive odd number is , n + 2 , since odd numbers have a difference of 2 between them.
from the given statement :
# n^2 + (n+2)^2 = 202#
expanding gives :
# n^2 + n^2 + 4n + 4 =202#
this is a quadratic equation so collect terms and equate to zero.
# 2n^2 + 4n -198 = 0 #
common factor of 2 :
# 2(n^2 + 2n - 99) = 0#
now consider factors of -99 which sum to +2. These are 11 and -9.
hence : 2(n + 11 )(n-9 ) = 0
(n + 11 ) = 0 or (n-9) = 0 which leads to n = -11 or n = 9
but n > 0 hence n = 9 and n+ 2 = 11
Always remember that
So,let the first number be
Then the second number will be
Now this is a Quadratic equation (in form
Luckily,we can factor it to
Now we have two values for
Now we need to find
So,at the end we conclude the if the first integer is