A few days ago
victoria 123

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
Anonymous

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.

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A few days ago
Jr Levesque
smalller odd number = 3

bigger odd number = 5

3(3)*2 = 2 + (5)*2

27=27

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A few days ago
Anonymous
There aren’t two consecutive odd numbers… they all have an even number separating them.
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A few days ago
Mikey B
this is a very easy one – its 3 and five – work on the info included in the question and you’ll see its correct.
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