SOLUTION: The product of 2 consecutive positive integers is 71 more than their sum. Find the integers. I have been reviewing my text for a couple of hours now. I cannot figure this one ou

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Question 339346: The product of 2 consecutive positive integers is 71 more than their sum. Find the integers.
I have been reviewing my text for a couple of hours now. I cannot figure this one out. Please, please, please help!!!!
Thank you!!
Answer by Kalmetam(43) About Me  (Show Source):
You can put this solution on YOUR website!
Hi! Try to make your own equation, then it'll be easy..
So two consecutive positive integers?
Let's say one integer is "x" and since they are consecutive, that means the 2nd is x+1.. right?
So let's try to make an equation:
The PRODUCT of two consecutive integers.. so that would be
x(x+1)
and the second part is 71 more than their sum..
how can you express that without using other variables?
Well, the two numbers are x and x+1 right? and the product is 71 more.. right?
So you'd get on that side.. 71+[x+(x+1)]
So in the end.. x(x+1)=71+[x+(x+1)]
Try simplifying the left side:
x+x^2=71+[x+(x+1)]
Now simplify the right side:
x+x^2=72+2x
Move over that x from the left to the right to simplify it more
x^2=72+x
Now square root both sides to cancel out the x^2
x=sqrt%2872%2Bx%29
then if you plot bothsides on a graph
+graph%28+300%2C+200%2C+-20%2C+20%2C+-15%2C+15%2C+x%2C+sqrt%2872%2Bx%29%29+

So apparently, the answer is x=9 (meaning that the second number (x+1) is 10)
Let's check:
9*10=90
9+10=19
and the product is 71 more than the sum, right?
90-19=71..
So our answers must be correct..
the numbers are 9 and 10