A few days ago
find two consecutive odd numbers?
such that three times the square of the smaller number is two more than the square of the larger number
Top 4 Answers
A few days ago
Favorite Answer
An even number = 2n
An odd number = 2n + 1
A consecutive odd number = 2n + 3
We need it such that
{3[(2n+1)^2]}-2= (2n + 3)^2
{3[4n^2+4n+1]}-2= (4n^2 + 12n + 9)
{12n^2+12n+3}-2= (4n^2 + 12n + 9)
12n^2+12n+1 = (4n^2 + 12n + 9)
8n^2 – 8 = 0
8n^2 = 8
n^2 = 1
n = 1
If n=1, then
{3[(2n+1)^2]}=27
(2n + 3)^2=25
Is 25 = 27-2? Yes.
Your two consecutive numbers are 25 and 27.
0
A few days ago
smalller odd number = 3
bigger odd number = 5
3(3)*2 = 2 + (5)*2
27=27
1
A few days ago
There aren’t two consecutive odd numbers… they all have an even number separating them.
1
A few days ago
this is a very easy one – its 3 and five – work on the info included in the question and you’ll see its correct.
1
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