Question 102885
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:
{{{x(x+2)+(x+1)=41}}}
first multiply x across (x+2)
{{{x^2+2x+(x+1)=41}}}
next multiply 1 across (x+1)
{{{x^2+2x+x+1=41}}}
now combine like terms
{{{x^2+3x+1=41}}}
now move 41 over by subtracting it from both sides
{{{x^2+3x+1-41=41-41}}}
{{{x^2+3x-40=0}}}
OK so now we have a quadratic equation in standard form. We can solve for x by factoring or by using the quadratic formula.
*[invoke quadratic "x", 1, 3, -40]
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
<b>Answer: The three consecutive positive numbers are 5, 6, and 7</b>
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