find the intergers

Question

The product of two consecutive integers is 13 less then 5 times their sum . Find the intergers 

Answer 1

Let two consecutive integers be x & x+1

Product of two consecutive integers = x(x+1), sum = (x+x+1)

 

x(x+1) +13 = 5(x+x+1)

x² +x+13=10x+5

 

x² +x+13-10x-5=0

 

x²-9x+8=0

(x-8)(x-1)=0

 

x= 8 , x+1 = 9 

 Check :LHS=72+ 13=85 

            RHS =5(8+9)=85                                                   

 

or x=1, x+1=2

Check: LHS=2+13=15

           RHS=5(1+2)=15

Two consecutive integers are 8,9   or 1,2

 

Two consecutive integers are 8,9

 

 

Answer 2

n(n+1)  = 5( n + n+1)-13

 

 

n(n+1) = 5(2n + 1) - 13

 

n^2 + n = 10n + 5 - 13

 

n^2 + n = 10n - 8

 

n^2 + n - 10n + 8 = 0

 

n^2 - 9n + 8 = 0

 

(  n    -   8 )( n   -      1   )=0

 

n-8 = 0 ---> n = 8

 

n - 1 = 0 ---> n = 1

 

So there are two answers:

8 x 9 = 72 = 85 - 13 = 5 * 17 - 13 = 5 * (8 + 9) - 13  <---- yes it works!!!

 

1 x 2 = 2 = 15 - 13 = 5*3 - 13 = 5 * ( 1 + 2) - 13