The product of two consecutive odd integers is 1 less than 6 times their sum. Find the two integers.

Question

anonymous · Answer

let a is any odd integer,
a*(a+2)=6*(a+a+2)-1
$$a ^{2}-10a-11=0$$
$$a ^{2}+a-11a-11=0$$
$$a(a+1)-11(a+1)=0$$
$$(a+1)(a-11)=0$$
a=-1,11

anonymous · Answer

my online work said -1 was correct, but I need 2 different sets of values for my problem.

anonymous · Answer

@KingD @tanya123 , Are y'all going to help oooooor...?

tanya123 · Answer

Sorry but I dont know!

tanya123 · Answer

and I think @hanzi has already helped you.../

anonymous · Answer

I need more answer, plus one was incorrect :/

anonymous · Answer

-1 and 11 are the answers

anonymous · Answer

@ganeshie8 , please help :(

anonymous · Answer

I need 2 more answers, I'll take a picture...

anonymous · Answer

attachment

anonymous · Answer

-1 and (-1+2)=1
so, one set of answers is -1,1
similarly other set is 11,(11+2)
11,13

tkhunny · Answer

Note: a and a+2 are not necessarily odd.  If you get a = 2, you have a problem.

If you REALLY want odd, perhaps 2k+1, 2k+3 for k an integer.  This may be better, as it truly defines odd values, but we are still relying a little on luck.  What if k isn't an integer in our final solution.

anonymous · Answer

Got it! Thank you ! :)