Find 2 consecutive odd integers

Question

"Find two consecutive odd integers such that the square of the first, added to 3 times the second, is 34."

 

I know it has something to do with creating a quadratic and factoring it, but I keep getting incorrect answers!!

Answer 1

An odd integer can be written as

 

2n+1

 

Where n is any integer.

 

The next odd integer starting from n would be

 

2n+3

 

Given this, we can write the rest of the equation like so:

 

(2n+1)^2+3*(2n+3) = 34

 

We can expand this equation to

 

4*n^2+10*n+10=34

 

Factoring this equation results in

 

4*(n+4)*(n-3/2)=0

 

So we have solutions at n = -4 and n = 3/2

 

Since -4 is the only integer, that will be our value of n.

 

Plugging back into the first equation, we confirm that 

 

(2n+1)^2+3*(2n+3) | n = -4

(2(-4)+1)^2+3*(2(-4)+3) = 34

 

So, our two consecutive odd integers will be 

 

(2n+1) = (2(-4)+1) = -7

 

and

 

(2n+3) = (2(-4)+3) = -5