Seudónimo Seudónimo
  • 21-07-2016
  • Mathematics
contestada

Find two consecutive counting numbers whose squares differ by 35.

Respuesta :

konrad509
konrad509 konrad509
  • 21-07-2016
[tex](n+1)^2-n^2=35\\ n^2+2n+1-n^2=35\\ 2n=34\\ n=17\\ n+1=18[/tex]

17 and 18
Answer Link
apologiabiology
apologiabiology apologiabiology
  • 21-07-2016
the numbers are n and n+1
the difference is 35
(n+1)^2-n^2=35
expand
n^2+2n+1-n^2=35
combine like terms
2n+1=35
minus 1 from oth sides
2n=34
divide by 2
n=17

the other number is 1 more

17 and 18
check
17^2=289
18^2=324
324-289=35


17 and 18
Answer Link

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