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Question 255865: solve:
The sum of the squares of two consecutive negative even integers is 100.
Find the integers.
Answer by unlockmath(1688)   (Show Source):
You can put this solution on YOUR website! Hello,
Let's set up -x will represent a negative integer and -x+2 will represent the next negative integer. Now we can set up our equation as:
(-x)^2 + (-x+2)^2=100 Rewrite this as:
x^2+x^2-4x+4=100 Subtract 100 from both sides to be:
x^2+x^2-4x-96=0 Combine like terms will result in:
2x^2-4x-96 Factor out a 2 will be:
2(x^2-2x-48)=0 This can be fractored as:
2(x-8)(x+6)=0 Now we know that:
x=8
x=-6
8 works as the answer so the negative integers are -8 and -6.
Square both of those and you'll see it adds up to 100.
Make sense?
RJ
Check out a new math book at:
www.math-unlock.com
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