SOLUTION: I am thinking of three consecutive positive numbers. If i multiply the first with the third and then add the second, the result is 41. Let x be the smallest number.

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Question 99076: I am thinking of three consecutive positive numbers. If i multiply the first with the third and then add the second, the result is 41. Let x be the smallest number.
Answer by rmromero(383)   (Show Source): You can put this solution on YOUR website!

I am thinking of three consecutive positive numbers. If i multiply the first with the third and then add the second, the result is 41. Let x be the smallest number.
What is asked in the problem?
find the 3 consecutive positive numbers.
Given:
the first is multiplied with the third and then add the second, the result is 41.
Representation:
Let x be the smallest number
x + 1 be the second number
x + 2 be the third/greatest number
Equation:
x (x+2) + x+1 = 41
x^2 + 2x + x + 1 = 41
x^2 + 3x + 1 = 41
x^2 + 3x - 40 = 0
(x + 8) (x - 5) = 0
x + 8 = 0 and x - 5 = 0
x = -8 x = 5
Since we are asked to find the three consecutive POSITIVE number
we reject -8.
x = 5
x + 1 = 5 + 1 = 6
x + 2 = 5 + 2 = 7
Therefore the three consecutive POSITIVE integers are 5, 6, and 7.

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