Question 166426
let x = smallest even number
let x + 2 = next smallest
let x + 4 = biggest.
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sum of the smallest 2 = product of the largest 2 / 4
x + x + 2 = [(x+2)*(x+4)]/4
simplify:
2*(x+1) = [(x+2)*(x+4)]/4
multiply both sides of equation by 4:
8*(x+1) = [(x+2)*(x+4)]
divide both sides of equation by (x+1):
8 =  [(x+2)*(x+4)]/(x+1)
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multiply numerator on right hand side of equation:
8 = (x^2 + 6*x + 8)/(x+1)
divide the numerator on right hand side of equation by the denominator:
8 = (x+1)*(x+5) + 3
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the 3 on the right hand side of the equation is the remainder of the division.
x+1 went into x^2 + 6*x + 8 (x+5) times with a remainder of 3.
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multiply factors on right hand side of equation together:
8 = x^2 + 6*x + 5 + 3
subtract 8 from both sides of the equation and combine like terms:
0 = x^2 + 6*x
complete the squares on the right hand side of the equation.
0 = (x+3)^2 - 9
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when you completed the squares, you got 9 more then what you started with so you had to subtract 9 to keep the original value of the equation intact:
x^2 + 6*x became (x+3)^2 after completing the squares.
(x+3)^2 = x^2 + 6*x + 9
since you have 9 more than you started with you needed to subtract 9 to keep the equation intact.
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add 9 to both sides of the equation.
9 = (x+3)^2
which is the same as:
(x+3)^2 = 9
take square root of both sides of the equation.
x+3 = +/- 3
subtract 3 from both sides of the equation:
x = -3 +/- 3
roots of equation are either:
x = 0
or:
x = -6
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answer should be either x = 0, or x = -6
try x = 0 first:
x = 0
x + 2 = 2
x + 4 = 4
add smaller numbers together:  2 + 0 = 2
multiply larger numbers together and divide by 4:  2*4 = 8/4 = 2
sum of the 2 smaller numbers equals product of 2 larger divided by 4.
x = 0 looks good.
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x = -6 didn't work.
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to get this solution, i needed to be able to divide polynomials, and i needed to be able to complete the squares.
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this is the first time i solved one of these this way (dividing polynomial with remainder and then completing the square).
i think it's valid.
not sure if this is how your instructor would solve this, but i did get an answer and it did work.
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fyi,
you were on the right track.
the solution was not readily apparent.
good luck.