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Question 413304: can you help me find three consecutive integers such that the square of twice the first is 30 more than three times the product of the second and third?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! find three consecutive integers such that the square of twice the first is
30 more than three times the product of the second and third?
:
Three consecutive integers, x, (x+1). (x+2)
:
(2x)^2 = 3(x+1)*(x+2) + 30
:
square 2x and FOIL
4x^2 = 3(x^2 + 3x + 2) + 30
4x^2 = 3x^2 + 9x + 6 + 30
:
Combine like terms on the left
4x^2 - 3x^2 - 9x - 36 = 0
:
x^2 - 9x - 36 = 9
Factors to
(x - 12)(x + 3) = 0
Two solutions
x = 12
and
x = -3
:
12, 13, 14 are our 3 integers
or
-3, -2, -1, could also be the 3 integers
:
:
Check both in the statement
(2*12)^2 = 3(13*14) + 30
576 = 546 + 30; confirms our solution of x=12
and
(2*-3)^2 = 3(-2 * -1) + 30
36 = 6 + 30; confirms the x = -3 solution
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