SOLUTION: Given three consecutive odd numbers such that the square of the second number is 192 less than the square of the third. Find those numbers.

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Question 634305: Given three consecutive odd numbers such that the square of the second number is 192 less than the square of the third. Find those numbers.
Found 2 solutions by graphmatics, MathTherapy:
Answer by graphmatics(170) About Me  (Show Source):
You can put this solution on YOUR website!
Let 2n+1 be the first odd integer, then 2n+3 and 2n+5 are the second and third consecutive odd integers
%282n%2B3%29%5E2%2B192=%282n%2B5%29%5E2
4%2An%5E2%2B12%2An%2B9%2B192=4%2An%5E2%2B20%2An%2B25
12%2An-20%2An=25-201
-8%2An=-176
n=22
The consecutive odd integers are 45,47 and 49.

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

Given three consecutive odd numbers such that the square of the second number is 192 less than the square of the third. Find those numbers.

Let the 1st of the integers be F

Then 2nd and 3rd are F + 2, and F + 4, respectively

Based on the given info, we see that: %28F+%2B+2%29%5E2+=+%28F+%2B+4%29%5E2+-+192

F%5E2+%2B+4F+%2B+4+=+F%5E2+%2B+8F+%2B+16+-+192

F%5E2+-+F%5E2+%2B+4F+-+8F+=+-+176+-+4

- 4F = - 180

F, or 1st integer = %28-+180%29%2F-+4, or 45

Therefore, the three consecutive odd integers are: highlight_green%2845_47_and_49%29

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