SOLUTION: I need help with this equation find three consecutive odd integers such 4 times the product of the sencond and third is 12 times greater that 20 times the sum of the first and sec

Question 118592: I need help with this equation find three consecutive odd integers such 4 times the product of the sencond and third is 12 times greater that 20 times the sum of the first and second
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
I think there is a typo here, it should read "12 greater" not "12 times greater"
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"find three consecutive odd integers"
x, (x+2), (x+4)
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" such 4 times the product of the second and third is 12 greater that 20 times the sum of the first and second"
4(x+2)(x+4) = (20(x + (x+2)) + 12
:
4(x^2 + 6x + 8) = 20(2x+2) + 12; FOILed (x+2)(x+4)
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4x^2 + 24x + 32 = 40x + 40 + 12
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4x^2 + 24x + 32 = 40x + 52
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4x^2 + 24x - 40x + 32 - 52 = 0; combine like terms on the left
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4x^2 -16x - 20 = 0; a quadratic equation
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x^2 - 4x - 5 = 0; simplified, divided equation by 4
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(x - 5)(x + 1) = 0; factored
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x = +5
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Our integers: 5, 7, 9
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Check solution in the statement:
"4 times the product of the second and third is 12 greater that 20 times the sum of the first and second"
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4*7*9 = 12 + 20(5 + 7)
252 = 12 + 20(12)
252 = 12 + 240; confirms our solution
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However, how about the other solution to our quadratic equation? x = -1
That would make the integers: -1, +1, +2, wouldn't it.
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Trying that in our statement:
"4 times the product of the second and third is 12 greater that 20 times the sum of the first and second"
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4 * +1 * +3 = 12 + 20(-1 +1)
12 = 12 + 20(0); confirms our other solution:
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So we have: 5, 7, 9; and -1, +1, +3; pretty neat, don't you think?
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