SOLUTION: (x+3)squared-(x+5)squared=-40

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Question 1132714: (x+3)squared-(x+5)squared=-40
Found 2 solutions by stanbon, greenestamps:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(x+3)squared-(x+5)squared=-40
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x^2+6x+9-(x^2+10x+25) = -40
-4x-16 = -40
x + 4 = 10
x = 6
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Cheers,
Stan H.
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Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


To simplify things, let's rewrite the equation as

%28x%2B5%29%5E2-%28x%2B3%29%5E2=+40

Note the solution by the other tutor, using formal algebra, is fine. But this problem can be solved mentally using the fact that the difference between the squares of consecutive integers is always an odd number, equal to 1 more than twice the smaller number:

%28n%2B1%29%5E2+-+n%5E2+=+2n%2B1

What we have is two numbers that differ by 2, and the difference of their squares is 40.

Since 40 = 19+21, that means the difference between the smaller integer n and the next integer n+1 is 19, and the difference between n+1 and the next integer n+2 is 21.

That in turn means the smaller integer is 9.

So x+3 is 9, making x = 6.

CHECK: (6+3)^2 - (6+5)^2 = 9^2-11^2 = 81-121 = -40