SOLUTION: find the second of the three consecutive positive even integers such that the product of the first and second is 64 less than the square of the third

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Question 390059: find the second of the three consecutive positive even integers such that the product of the first and second is 64 less than the square of the third
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
Let x, (x+2),(x+4) represent the three consecutive positive even integers
Question States***
x(x+2) = (x+4)^2 - 64
solving for x
x^2 + 2x = x^2 + 8x + 16 - 64
48 = 6x
x = 8, the first of the three. The Second Integer is 10
CHECKING our Answer*****
8*10 = 144 - 64

Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!

given:
the three consecutive positive even integers
Assign variables :
Let be the first consecutive positive even integers
be the second consecutive positive even integers
be the third consecutive positive even integers
the product of the first and second is less than the square of the third

...or

.....................the first consecutive positive even integer

..............the second consecutive positive even integer

...............the third consecutive positive even integer

check: