Question

Please help!!!!

Answer 1

Find two consecutive odd integers such that their product is 11 more than 2 times their sum.

Answer 2

hint:

let n = the first odd integer
let n + 2 = the second odd integer

any ideas where to go from here?

Answer 3

13??

Answer 4

not quite

Answer 5

hmm. Okay how would I do this.

Answer 6

"product" means multiply so...

(n)(n+2)

"11 more" means + 11 so...

(n)(n+2) + 11

"2 times their sum" means 2(n + n + 2)

so put everything together

(n)(n+2) + 11 = 2(n + n + 2)

Answer 7

then solve for n

Answer 8

so should i start to simplify it?

Answer 9

yes

Answer 10

start with the left side

(n)(n+2) = ?

Answer 11

4n?

Answer 12

or n^2+2n

Answer 13

the second one

Answer 14

2(n + n + 2) = ?

Answer 15

4n+2

Answer 16

almost

Answer 17

4n + 4

Answer 18

opps okay

Answer 19

n^2+2n+11=4n+4

Answer 20

so now we get:

n^2 + 2n + 11 = 4n + 4

keep simplifying

Answer 21

n^2=2n+15

Answer 22

almost

Answer 23

n^2 = 2n - 7

Answer 24

okay

Answer 25

hold on a sec let me check my work

Answer 26

(made a typo earlier)

n^2 + 2n = 4n + 4 + 11

which gives us

n^2 - 2n - 15

that factors to

(n-5)(n+3)

so one of our integers is either n = 5 or n = -3

to save some time, I tested it and it's n = -3, meaning that the consecutive integer is -1

so your answers are n = -3 and -1