the produtof two consecutive positive integers is 1 more than theirsum. find the two integers.

Question

the produtof two consecutive positive integers is 1 more than theirsum. find the two integers.

 

Answer 1

Hi April,

 

Let x = the first number

 

x+1 = the second number

 

now create an equation that matches the sentence

 

x(x+1)  (product of two consecutive integers)

 

x+(x+1)+1  (one more than the sum of the two integers)

 

set them equal

 

x(x+1) = x+(x+1)+1

 

x²+x = 2x +2

 

x² -x -2 = 0

 

factor

 

(x-2)(x+1)=0

 

x+1=0

x=-1    Does not work we need positivie integers

 

x-2=0

 

x = 2

 

and

 

x+1=3

 

Product 2*3 =6

 

Sum 2+3 = 5

 

Product is 1 more than sum

 

It all works

 

Answer 2

First, let us set up the equation.  A number and its consecutive are n and n+1.  So, our equation looks like this:

n(n+1) = n + n+1 + 1

Simplify:

n^2 + n = 2n + 2

Subtract (2n+2) from both sides:

n^2 + n - 2n -2 =0

Simplify:

n^2 -n -2 = 0

Factor:

(n-2)(n+1) = 0

The product will be 0 when one factor is 0, so either n - 2 = 0  or  n + 1 = 0.

This means either n = 2   or  n = -1.  We disregard this second possibility, because we are given that we are working with positive numbers.

So, the numbers are n=2, and n + 1 = 2+1 = 3:  2 and 3.

We check:  2 * 3 does indeed equal 2 + 3 + 1.

:)