SOLUTION: find three consecutive integers such that the product of the first two plus the product of the last two is 8

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Question 1119463: find three consecutive integers such that the product of the first two plus the product of the last two is 8
Found 3 solutions by math_helper, ankor@dixie-net.com, MathTherapy:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Let m=the smallest of the three consecutive integers

Then +%28m%29%28m%2B1%29+%2B+%28m%2B1%29%28m%2B2%29+=+8+

Which reduces to +m%5E2+%2B+2m+-+3+=+0+
Solutions are +m+=+%28-2+%2B-+sqrt%2816%29%29%2F2+
Potential solutions are m = -3 and m = 1.


Check them:
m=-3: (-3)(-2) + (-2)(-1) = 6+2 = 8 ( ++m+=+-3+ is a solution, and the corresponding integers are +highlight%28matrix%281%2C4%2C+%22-3%2C+%22%2C+%22-2%2C+%22%2C+%22and+%22%2C+%22-1%22%29%29+ )


m=1: (1)(2) + (2)(3) = 2 + 6 = 8 ( ++m+=+1+ is a solution, and the corresponding integers are +highlight%28matrix%281%2C4%2C+%221%2C+%22%2C+%222%2C+%22%2C+%22and+%22%2C+%223%22%29%29+ )
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Tutor ankor@dixie-net made this mistake:
The line
a^2 + a + a^2 + 2a + 1 + 2 = 8
should be +a%5E2+%2B+a+%2B+a%5E2+%2B+2a+%2B+green%28a%29+%2B+2+=+8+ that is why he did not find the 2nd solution.

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@MathTherapy - thanks for sharing that most efficient solution!

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive integers
a, a+1, a+2
such that the product of the first two plus the product of the last two is 8
a*(a+1) + (a+1)(a+2) = 8
a^2 + a + a^2 + 2a + 1 + 2 = 8
2a^2 + 3a + 3 - 8 = 0
2a^2 + 3a - 5 = 0
Factors to
(2a+5)(a-1) = 0
only one integer solution
a = 1
:
the integers 1, 2, 3
:
:
check
(1*2) + (2*3) = 8

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive integers such that the product of the first two plus the product of the last two is 8
One of the easiest methods to solve this is to first let the 2nd integer be S

Then 1st = S - 1, and 3rd: S + 1

From what was given, we can say that: (S - 1)S + S(S + 1) = 8

matrix%281%2C3%2C+S%5E2+-+S+%2B+S%5E2+%2B+S%2C+%22=%22%2C+8%29 

matrix%281%2C3%2C+2S%5E2%2C+%22=%22%2C+8%29 =====> matrix%281%2C3%2C+S%5E2%2C+%22=%22%2C+4%29

Therefore, 2nd integer, or 



If the 2nd = 2, then the 1st and 3rd = S - 1, or 2 - 1, or 1, and S + 1, or 2 + 1, or 3. Integers are therefore: highlight_green%28matrix%281%2C4%2C+%221%2C%22%2C+%222%2C%22%2C+and%2C+%223%22%29%29



If the 2nd = - 2, then the 1st and 3rd = S - 1, or - 2 - 1, or - 3, and S + 1, or - 2 + 1, or - 1. Integers are therefore: highlight_green%28matrix%281%2C4%2C+%22-+3%2C%22%2C+%22-+2%2C%22%2C+and%2C+%22-+1%22%29%29

You're most welcome, @math_helper. Was more than happy to share!