SOLUTION: two consecutive postive integers sum of squares is 85

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Question 53317: two consecutive postive integers sum of squares is 85
Found 2 solutions by stanbon, EduDragon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consecutive numbers be "x" and "x+1".
EQUATION:
x^2 + (x+1)^2= 85
x^2+ x^2+2x+1 = 85
2x^2+2x-84=0
x^2+x-42=0
(x+7)(x-6)=0
x=6 or x=-7
The consecutive numbers are -7, -6
or 6,7
Cheers,
Stan H.

Answer by EduDragon(8) About Me  (Show Source):
You can put this solution on YOUR website!
Two consecutive numbers would be n and %28n%2B1%29.
Their squares would be n%5E2 and %28n%2B1%29%5E2.
The squares sum would is equal to 85.
Set the equation so that it looks like this: n%5E2+%2B+%28n%2B1%29%5E2+=+85.
Now solve for n.
First combine all like terms by multiplying out %28n%2B1%29%5E2. When you do you get n%5E2+%2B+2n+%2B+1. When you combine all like terms you get 2n%5E2+%2B+2n+%2B+1+=+85.
Now set the whole equation equal to zero by subtracting 85 from both sides to get: 2n%5E2+%2B+2n-84=0.
Second, factor out a 2 from everything; that leaves you with 2%28n%5E2%2Bn-42%29=0.
Factor the parenthesis to get 2%28n%2B7%29%28n-6%29=0.
Now set each set of parenthesis to zero giving you n%2B7=0 and n-6=0.
Solve for n in each case. Do that by either adding or subtracting the appropriate number. This leaves you with n=-7 and n=6.
Since the problem wants positive integers you can rule out the -7 leaving the answer to be 6 and, therefore, 7 since you want consecutive integers.