SOLUTION: the problem says find three consecutive even intergers such that the sum of the smaller two is one fourth the product of the second and third interger. So far i got x+x+2=1/4[(x+2)

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Question 166426: the problem says find three consecutive even intergers such that the sum of the smaller two is one fourth the product of the second and third interger. So far i got x+x+2=1/4[(x+2)(x+4)] but after that im lost. Can you help me?
Found 2 solutions by MRperkins, gonzo:
Answer by MRperkins(300) (Show Source):
You can put this solution on YOUR website!
you are absolutely correct in setting up your equation. Good job!
Now you just need to do the math to find out what x is.
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Left side
x+x+2=2+2x
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Right side
you need to foil (x+2)(x+4) and then distribute the throughout all the terms.

=

combine like terms

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distribute

so you have
left side = right side

get rid of the 4 in the denominator by multiplying everything on the left and right side by 4.

get everything on one side
subtract (8+8x) from each side *make sure you use parenthesis() so you subtract both terms

distribute the negative

combine like terms

.
factor out an x

x=0;
x=2
either one of these will work but since 0 is neither even or odd you can not use x=0 therefore x=2.
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So, your three consecutive even integers are x, x+2, x+4 or 2,4,6
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Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
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.