Question 534044
The product of second and third  of 3 consecutive integers is 3 more than 3 times the second integer. find the 3 integers


Let the 1st integer be F


Then the other 2 are:  F + 1, and F + 2


We have: (F + 1)(F + 2) = 3(F + 1) + 3


{{{F^2 + 3F + 2 = 3F + 3 + 3}}}


{{{F^2 + 3F - 3F = 6 - 2}}}


{{{F^2 = 4}}} 


F, or 1st integer = ± 2


Therefore, the consecutive integers are either:{{{highlight_green(2_3_and_4)}}}, or {{{highlight_green(-2_-1_and_0)}}}


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Check
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Integers are: 2, 3, & 4
Product of 2nd and 3rd: (3 * 4) = 3(3) + 3 ---- 12 = 9 + 3 ---- 12 = 12 (TRUE)


Integers are: - 2, - 1, & 0
Product of 2nd and 3rd: (- 1 * 0) = 3(- 1) + 3 ---- 0 = - 3 + 3 ---- 0 = 0 (TRUE)


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