# The sum of the square of two consecutive positive odd integers is 202, how do you find the integers?

##### 2 Answers

#### Answer:

9 , 11

#### Explanation:

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

Then,

Use formula

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

If

Then,

And if

Then,

So,at the end we conclude the if the first integer is