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
----
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
=====================
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

  

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