Question

find 3 consecutive odd integers such that the square of the first decreased by the square of the third is equal to 10 less than twice the square of the second

Answer 1

\[x^2-(x+3)^2=2(x+2)^2-10\]

Answer 2

distribute the brackets first

Answer 3

This is what I got so far, but I know something is wrong

Answer 4

the intengers should look like these
x x+2 x+4..........because you need odd intengers

Answer 5

FYI I translated the problem as an equation but something is wrong. HELP!!!!

Answer 6

x
x+2
x+4
\[x^2-(x+4)^2=(x+2)^2-10\]

Answer 7

instead of x+3 you should have x+4, just like @Mertsj wrote

Answer 8

lol I got it

Answer 9

Thanks for the help guys!!

Answer 10

ok. np :)

Answer 11

yw