math help pleaseee?

Question

The positive difference between the squares of two consecutive odd integers is 128. What is the product of the two integers?
show work if possible :]

Answer 1

Let the second odd number be (x+2) and the first one x. Note that the difference between two consecutive odd or even numbers is 2, thus the +2.

Now the below formula is lined:

(x+2)^2 - x^2 = 128

This formula will help us expand the first binomial expression:

(a+b)^2 = a^2 + 2ab + b^2

(x+2)^2 - x^2 = 128
x^2 + 4x + 4 - x^2 = 128
4x + 4 = 128
4x = 124
x = 31

Therefore, the first odd number is 31 and the second is 33. Now we want to calculate their product so let's multiply them:

31*33 = 1023

Good luck.

Answer 2

You want consecutive odd integers the difference of whose squares is 128

You want consecutive odd integers. So let one be 2n-1 and the other be 2n+1

(2n-1)²=4n²-4n+1
(2n+1)²=4n²+4n+1
Subtracting the smaller from the larger you get 8n

We’re told that the difference is 128 so
8n=128
n=16

2n-1=(2)(16)-1=32-1=31
2n+1=32+1=33

31²=961
33²=1089
1089-961=128
Checks out.

So, 31*33 = 32²-1=1023 Did that in my head. (32+1)(32-1)=32² - 1 = 1024-1

Answer 3

two odd integers can be represented by n and (n+2)

difference means subtract

(n+2)² - n² = 128
n² + 4n + 4 - n² =128
4n + 4 = 128
4n = 124
n = 124/4 = 31

numbers are 31 and 33

product of 31 and 33 is 1023

=]

..

Answer 4

Close. Odd integers are represented by
2n+1 so that two consecutive odd integers would be 2n+1 and 2n+3. Now proceed as you were to find
(2n+3)² - (2n+1)² = 128 => n = 15 so
2n + 1 = 31
2n + 3 = 33

HTH

Doug

Answer 5

(x+2)^2 - x^2 = 128
x^2 + 4x + 4 - x^2 = 128
4x + 4 = 128
4x = 124
x = 124/4
x = 31

The two consecutive odd integers are 31 and 33. Their product is 1023.