SOLUTION: Two consecutive positive integers such that the square of the second integer added to 4 times the first is equal to 28

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Question 602468: Two consecutive positive integers such that the square of the second integer added to 4 times the first is equal to 28
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Two consecutive positive integers such that the square of the second integer added to 4 times the first is equal to 28
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1st: x
2nd: x+1
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Equation:
(x+1)^2 + 4x = 28
x^2+2x+1 + 4x = 28
x^2+6x-27 = 0
Factor:
(x+9)(x-3) = 0
Positive solution:
1st: x = 3
2nd: x+1 = 4
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Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
Let x and (x+1) represent the Two consecutive positive integers
Question states***

Solving for x
x^2 + 2x + 1 + 4x = 28
x^2 + 6x -27 = 0
(x-3)(x+9)= 0
x = 3
the Two consecutive positive integers are 3,4
CHECKING our Answer***
16 + 12 = 28