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Word problem.
- Thread starter Rainbow
- Start date
D
Deleted member 4993
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Find two consecutive odd numbers such that the sum of their squares is 130.
I found out that it was 9(squared) + 7(squared), but I have to represent this as an quadratic equation and be able to solve it for 7 and 9.
Two consecutive odd numbers can be represented as (2*n-1) & (2*n+1).
now continue....
Please share your work with us .
If you are stuck at the beginning tell us and we'll start with the definitions e.g. cost, equity & debt.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...217#post322217
Since the squares are added (not multiplied), isn't it much easier to use n and n+2 ?
Consider this alternative:
Let n - 1 = the smaller odd integer
Let n + 1 = the larger odd integer
\(\displaystyle (n - 1)^2 \ + \ (n + 1)^2 \ = \ 130\)
\(\displaystyle n^2 - 2n + 1 \ + \ n^2 + 2n + 1 \ = \ 130 \)
\(\displaystyle 2n^2 + 2 \ = \ 130\)
\(\displaystyle n^2 + 1 \ = \ 65\)
\(\displaystyle n^2 \ = \ 64\)
Continue . . .
Rainbow, 7 and 9 are not the only consecutive odd numbers such that the sum of their squares is 130.
-9 and -7 are also consecutive odd numbers with the same property.