Question 981156: Find two consecutive positive integers such that the square of the second integer added to 5 times the first is equal to 229.omarroberts0 Found 2 solutions by Alan3354, LinnW:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find two consecutive positive integers such that the square of the second integer added to 5 times the first is equal to 229.
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Use x and x+1
(x+1)^2 + 5x = 229
x^2 + 7x - 228 = 0
(x-12)*(x+19) = 0
x = 12
--> 12 & 13
You can put this solution on YOUR website! Let x equal the first number
x + 2 equals the second number
The square of the second is (x + 2)^2
5 times the first number is 5(x)
(x + 2)^2 + 5(x) = 229
x^2 + 4x + 4 + 5x = 229
x^2 + 9x + 4 = 229
add -4 to each side
x^2 + 9x = 225
add -225 to each side
x^2 + 9x - 225 = 0
It turns out that there is not an integer solution to the equation
so there is not a solution to the problem as stated.
