The product of two consecutive even integers is 48. Find the two integers.

Question

1) The product of two consecutive even integers is 48. Find the two integers.

 

2) The product of two consecutive even integers is 22 less than 11 times their sum. Find the two integers.

 

3) The product of two consecutive odd integers is 1 less than 3 times their sum. Find the two integers. 

 

Answer 1

1)  Two consecutive even integers are 2n and 2(n+1), where n = any integer.

 

2n*2(n+1) = 48                        (The product of the two integers is 48)

 

4n²+4n-48=0                            (Multiply out, subtract 48 from both sides)

 

n²+n-12=0                                (Divide both sides by 4)

 

(n+4)(n-3)=0                             (Factor the quadratic)

 

n= -4 and +3

 

For n= -4, 2n = -8 and 2(n+1) = -6.  (-8)(-6) = 48

 

For n=3, 2n = 6 and 2(n+1) = 8.  (8)(6) = 48

 

There are two answers: -8 and -6, and 6 and 8.

 

2)  This one works the same way as #1. The starting equation is:

2n*2(n+1) = 11(2n + 2n + 2) - 22  (The product of the integers is 22 less than 11 times their sum)

 

Multiply out, combine like terms, and solve the quadratic.  Remember, the answers are 2n and 2(n+1), not n.

 

3)  This one is the same as the other two, except two consecutive odd integers are (2n+1) and (2n+3) for any integer n.  The starting equation is:

 

(2n+1)(2n+3) = 3(2n+1+2n+3) - 1  (The product of the two integers is 1 less than 3 times their sum)

 

Multiply out, combine like terms, and solve the quadratic.  Remember, the answers are 2n+1 and 2n+3, not n.