Question 417548
Find three consecutive even integers such taht the square of the sum of the first and second integers is equal to twice the third integer.
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Let x = first of three consecutive even integers
then
x+2 = second integer
x+4 = third integer
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(x + x+2)^2 =2(x+4)
(2x+2)^2 =2x+8
(2x+2)(2x+2) =2x+8
4x^2 + 8x + 4 = 2x + 8
4x^2 + 6x + 4 = 8
4x^2 + 6x - 4 = 0
2x^2 + 3x - 2 = 0
(2x-1)(x+2) = 0
x = {1/2, -2}
since we were looking for an integer we can throw out the 1/2 leaving:
x = -2
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solution: -2, 0, 2