# The product of two consecutive odd integers is 1 less than twice their sum. How do you find the integers?

Feb 9, 2016

3 and 5

#### Explanation:

let an odd integer be n then the next odd integer will be (n+2), since there is a difference of 2 between odd integers.

their product is n(n+2)

1 less than twice their sum is 2( n + n + 2 ) -1 = 2(2n + 2 ) -1

hence n(n+2) = 2(2n + 2 ) - 1 = 4n + 4 - 1 = 4n + 3

distribute brackets : ${n}^{2} + 2 n = 4 n + 3$

this is a quadratic. To solve collect terms to left side and equate to zero.

${n}^{2} - 2 n - 3 = 0$

factorising to obtain : (n-3)(n+1) = 0

hence (n-3 ) =0 → n = 3 or (n+1) = 0 → n = -1

but n ≠ -1 , hence n = 3 and n + 2 = 5

check : product # = 3 xx 5 = 15

1 less than twice their sum : 2(3+5) -1 =$2 \times 8 - 1 = 16 - 1 = 15$