Yahoo Answers: Answers and Comments for Find three consecutive odd integers such that 3 times the product of the first and third exceeds the product of the first and second by 28. ? [Mathematics]
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From atva
enGB
Mon, 18 Nov 2019 21:53:54 +0000
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Yahoo Answers: Answers and Comments for Find three consecutive odd integers such that 3 times the product of the first and third exceeds the product of the first and second by 28. ? [Mathematics]
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From no sea nabo: (2n1)(2n+3)=3x
(2n1)(2n+1)=x+28
n=3 x=7
(...
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Mon, 18 Nov 2019 22:26:13 +0000
(2n1)(2n+3)=3x
(2n1)(2n+1)=x+28
n=3 x=7
(7)(3)=3(7)
(7)(5)=(7)+28
LQQD
(3),(5),(7)

From llaffer: Three consecutive odd integers. I'll use:...
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Mon, 18 Nov 2019 22:14:13 +0000
Three consecutive odd integers. I'll use:
(x  2), x, and (x + 2)
So as long as "x" is odd, the other two will be as well.
Three times the product of the first and third:
3(x  2)(x + 2)
Exceeds the product of the first and second by 28:
3(x  2)(x + 2) = x(x  2) + 28
We can simplify both halves and solve as a quadratic:
3(x²  4) = x²  2x + 28
3x²  12 = x²  2x + 28
2x² + 2x  40 = 0
x² + x  20 = 0
(x + 5)(x  4) = 0
x = 4 and 5
Since we only want odd values for x we can throw out the 4 to get our answer to be:
7, 5, and 3

From Philip: Let 3 consecutive odd integers be 2n3, 2n1, ...
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Thu, 21 Nov 2019 03:15:47 +0000
Let 3 consecutive odd integers be 2n3, 2n1, 2n+1...(1).
Then 3(2n3)(2n+1) = (2n3)(2n1) + 28, ie., (2n3)(6n+32n+1) = 28, ie.(2n3)(n+1) = 7, ie., 2n^2 n 10 = 0, ie.,
(2n5)(n+2) = 0. Clearly, (1) holds only when n = 2. Then
the 3 consecutive odd integers are 7, 5, 3.
Check: 3(2n3)(2n+1) = 3(7)(3) = 63
(2n3)(2n1) + 28 = (7)(5) +28 = 35+28 = 63.

From la console: a = 2n  1 ← an odd number cannot be divided b...
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Wed, 20 Nov 2019 12:35:23 +0000
a = 2n  1 ← an odd number cannot be divided by 2
b = a + 2
c = b + 2
3 times the product of the first and third exceeds the product of the first and second by 28.
3ac = ab + 28
3.(2n  1).(b + 2) = (2n  1).(a + 2) + 28
3.(2n  1).(a + 2 + 2) = (2n  1).(2n  1 + 2) + 28
3.(2n  1).(2n  1 + 2 + 2) = (2n  1).(2n  1 + 2) + 28
3.(2n  1).(2n + 3) = (2n  1).(2n + 1) + 28
3.(4n² + 6n  2n  3) = (4n² + 2n  2n  1) + 28
3.(4n² + 4n  3) = (4n²  1) + 28
12n² + 12n  9 = 4n²  1 + 28
8n² + 12n = 36
n² + (12/8).n = 36/8
n² + (3/2).n = 9/2
n² + (3/2).n + (3/4)² = (9/2) + (3/4)²
n² + (3/2).n + (3/4)² = (9/2) + (9/16)
[n + (3/4)]² = 81/16
n + (3/4) = ± 9/4
n =  (3/4) ± (9/4)
n = ( 3 ± 9)/4
First case: n = ( 3 + 9)/4 = 6/4 = 3/2 ← no possible
Second case: n = ( 3  9)/4 =  12/4 =  3 ← ok
a =  7
b =  5
c =  3

From Como: :
Let integers be 2x + 1 , 2x + 3 , 2x +...
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Tue, 19 Nov 2019 15:03:43 +0000
:
Let integers be 2x + 1 , 2x + 3 , 2x + 5
3 ( 2x + 1) ( 2x + 5 ) = ( 2x + 1 ) (2x + 3) + 28
3 ( 4x² + 12x + 5 ) = 4x² + 8x + 31
8x² + 28x  16 = 0
2x² + 7x  4 = 0
( 2x  1 ) ( x + 4 ) = 0
x =  4 is an integer value
The three integers are ( 7) , (5) , (3)

From Krishnamurthy: Three consecutive odd integers are such that
...
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Mon, 18 Nov 2019 22:37:05 +0000
Three consecutive odd integers are such that
3 times the product of the first and third
exceeds the product of the first and second by 28.
Let (2n – 1), (2n + 1), (2n + 3) be 3 consecutive odd integers.
We have 3(2n – 1)(2n + 3) – (2n – 1)(2n + 1) = 28
12n^2 + 12n – 9 – (4n^2 – 1) – 28 = 0
8n^2 + 12n – 36 = 0
2n^2 + 3n – 9 = 0
(2n – 3)(n + 3) = 0
n = –3 or n = 3/2
The three consecutive odd integers are 7, 5, 3.

From Mercy: example:
https://www.algebra.com/algebra/home...
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Mon, 18 Nov 2019 22:17:01 +0000
example:
https://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.338915.html

From Anonymous: They are both 8 and 12.
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Mon, 18 Nov 2019 22:08:51 +0000
They are both 8 and 12.

From Greg: You have a mistake in your question.
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Mon, 18 Nov 2019 22:10:24 +0000
You have a mistake in your question.