Question

Three consecutive odd integers are such that the square of the second is 96 less than the square of the third. Find the three integers.

This hurts me!

Answer 1

assume x is odd
what is the next odd number after x
x+1 is even
so x+2 is odd
next odd number is x+4
so we have
(x+2)^2=(x+4)^2-96

Answer 2

(x+2)^2=(x+2)(x+2)=x^2+2x+2x+4=x^2+4x+4
(x+4)^2=(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16

---------
x^2+4x+4=x^2+8x+16-96
x^2+4x+4=x^2+8x-80
x^2-x^2+4x-8x+4+80=0
0-4x+80=0
-4x=-80
x=20
x+2=22
x+4=24

Answer 3

21,23,25

Answer 4

I would let x be the 2nd number

then solve \[x^2+96=(x+2)^2\]

Answer 5

oops
x^2+4x+4=x^2+8x-80
x^2-x^2+4x-8x=-80-4
-4x=-84
x=21
so x+2=23
and
x+4=25

Answer 6

i forgot to add 4 and 80 :(

Answer 7

i think im dyslexic

Answer 8

not good

Answer 9

i don't think it ever helped anyone to be dyslexic

Answer 10

no

Answer 11

Awesome!