Please look- i don't know what i'm doing wrong? two consecutive odd intergers?

Question

Thanks for helping with my math questions. I'm helping my bf with his homework, but I took algebra 2 two years ago and my brain is foggy...

This problem says:

Find two consecutive odd intergers such that the square of the first increased by the second is 32.

Logically I know that 5^2 is 25, and 25 plus 7 is 32. 5 and 7 are consecutive and odd....

here's where my problem comes in.

My boyfriend has to have an equation and putting this into an equation I would get:

n^2 + (n + 2) = 32
n^2 + n - 30 = 0
(n+6) (n-5)= 0

This means that according to equation form, the numbers are 5 and 6. However 6 is not odd and 5^2 + 6 is 31...


Am I missing something? Did I write the equation wrong?

Answer 1

your numbers are N and N + 2. You solved for N and got 5 and -6.

The only odd integer out of the possible answers is 5.
N = 5,
N + 2 = 7.

You can throw out n = -6 because it doesn't fit the criteria of the question.

Answer 2

Set up your variables as n and n + 2 since odd numbers are two apart.
Your equation is n^2 + n + 2 = 32
n^2 + n - 30 = 0
(n+ 6) (n - 5) = 0
n = {-6, 5}

Since you're only interested in odd integers, you won't use the -6.
So n = 5 and n + 2 = 7.

Answer 3

According to (n+6)(n-5)=0, the smaller of the two consecutive odd integers (which is what you made n equal to) is either -6 or 5. -6 doesn't fit the parameters, so it must be 5. Then the larger of the consecutive odd integers is the next odd up from 5, 7.

Answer 4

somebody else used this good judgment already: enable x be the smaller of those to unusual integers then x+2 is the bigger. x+(x+2) =ninety two 2x+2 = ninety two 2x=ninety so x = 40 5 and x+2 = 40 seven. Now my question is, are there 2 consecutive even integers whose sum is ninety two? Why or why no longer?

Answer 5

try this:
n^2+[(n^2)+2]=32