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. This is ho

Question 102885: 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.
This is how far I have made, but I don't think I'm doing it right.
x
x+1
x+2
x+x+1+x+2=41
3x+3=41
3x+3-3=41-3
3x=38
x=38/3
Answer by doukungfoo(195) (Show Source): You can put this solution on YOUR website!
This part of what you did is correct:
x
x+1
x+2
This is where you made a mistake:
x+x+1+x+2=41
The problem states:
If I multiply the first with the third and then add the second, the result is 41
So your equation should look like this:

first multiply x across (x+2)

next multiply 1 across (x+1)

now combine like terms

now move 41 over by subtracting it from both sides

OK so now we have a quadratic equation in standard form. We can solve for x by factoring or by using the quadratic formula.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=169 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5, -8. Here's your graph:

So the possible solutions for x are -8 and 5
However the problem states:
I am thinking of three consecutive positive numbers.
So we can throw out -8 as an extraneous solution.
Ok so the smallest number is 5
the next number is 5+1 which is 6
and the last number is 5+2 which is 7
Answer: The three consecutive positive numbers are 5, 6, and 7
Check by mulitiplying the first number by the third number and adding the second number. The result should be 41.
5 * 7 + 6 = 41
35 + 6 = 41
41 = 41

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