SOLUTION: The sum of the squares of two consecutive negative odd integers is 202. Find the integers.

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Question 1061153: The sum of the squares of two consecutive negative odd integers is 202. Find the integers.
Found 2 solutions by math_helper, Boreal:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

Let x = first number
then x+2 = 2nd number
+x%5E2+%2B+%28x%2B2%29%5E2+=+202+
++x%5E2+%2B+%28x%5E2+%2B+4x+%2B+4%29+=+202+
+++2x%5E2+%2B+4x+%2B+4+=+202+
++++x%5E2+%2B+2x+%2B+2+=+101+
+++x%5E2+%2B+2x+-+99+=+0+
++++%28x-9%29%28x%2B11%29+=+0+

x=9 or x=-11
Problem statement said the numbers are negative, so x = -11 and x+2 = -9.
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Answer: -11 and -9
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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
integer one is x integer two is x+2
Their squares are x^2 and (x+2)^2
The sum is 2x^2+4x+4=202
2x^2+4x+2=202
2x^2+4x-198=0
x^2+2x-99=0
(x+11)(x-9)=0
9 and -11 work
but -9 and -11 also work. Those are the integers, and the sum of their squares is 81+121=202.