Question

(PLEASE HELP)
Three consecutive odd integers are such that the square of the third integer I s9 less than the sum of the squares of the first two. One solution is -3, -1, and 1. Find three other consecutive odd integers that also satisfy the given conditions.
( I keep getting the wrong answer)

Answer 1

HI!!

Answer 2

Hello.

Answer 3

i assume you solved
\[n^2+(n+2)^2=(n+4)^2+9\]

Answer 4

Honestly I don't know since I don't understand what exactly is wrong

Answer 5

ok lets go from the beginning
first number is \(n\) then the second would be \(n+2\) and the third would be \(n+4\) right?

Answer 6

yes

Answer 7

so sum of squares of the first two are
\[n^2+(n+2)^2\] and 9 more than the third is \((n+4)^2+9\)

Answer 8

you get this quadratic equation \[n^2+(n+2)^2=(n+4)^2+9\] which will have two solutions

Answer 9

one is \(-3\) so \(-3,-1,1\) is one answer

Answer 10

the other is \(7\) so \(7,9,11\) we can solve it if you like

Answer 11

yes please, so -3 and 7 are two of the answers?

Answer 12

yeah \[n^2+(n+2)^2=(n+4)^2+9\] expand get
\[n^2+n^2+4n+4=n^2+8n+16\]

Answer 13

oops forgot the 9 !

Answer 14

\[n^2+n^2+4n+4=n^2+8n+16+9\] thats better

Answer 15

then put all on one side get
\[n^2-4n-21=0\] factor and you get your two solutions

Answer 16

so there would be ONLY two solutions right?

Answer 17

two solutions for the first number \(n\)

Answer 18

(x-3)(x+7) would turn to 3,-7?

Answer 19

yeah it would but to factor correctly it is
\[(x-7)(x+3)=0\]

Answer 20

ohh alright I got it from here thank you very much.