SOLUTION: The sum of the squares of two consecutive negative integers is 61. Find the smaller of the two integers.

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Question 399916: The sum of the squares of two consecutive negative integers is 61. Find the smaller of the two integers.
Found 2 solutions by Alan3354, Tatiana_Stebko:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
61/2 = 30.5
sqrt(30.5) =~ 5.5, the middle
--> 5 & 6
Also -6 & -5

Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
If the first integer is x, then the second is (x+1)(because they are the two consecutive integers)
the sun of their squares is x%5E2%2B%28x%2B1%29%5E2
From problem's text the sun of their squares is 61,
so we have equation:
x%5E2%2B%28x%2B1%29%5E2=61
(formula %28a%2Bb%29%5E2=a%5E2%2B2ab%2Bb%5E2, so %28x%2B1%29%5E2=x%5E2%2B2x%2B1
x%5E2%2Bx%5E2%2B2x%2B1=61
2x%5E2%2B2x%2B1-61=0
2x%5E2%2B2x-60=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A2%2A%28-60%29+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+484%29%29%2F4+
x+=+%28-2+%2B22%29%2F4+ or x+=+%28-2+-22%29%2F4+
x+=+5 or x+=+-6
x+=+5 extraneous root because in problem negative integers
So x=-6 is the first integer
the second is (x+1)=-6+1=-5
Answer: The smaller integer is -6