Find two positive consecutive odd integers such that the square of the larger integer is one less than twice the square of the small integer. (algebra 1)

5 and 7

Step-by-step explanation:

Set up an equation based on the given information

Two consecutive odd integers can be written as

n, and (n + 2)      where n is one odd integer and n + 2 is the next odd integer

***If 3 is one integer, than 3 + 2 = 5 is the next odd integer***

It tells us that:  "the square of the larger integer is one less than twice the square of the small integer', so

(n + 2)² represents the square of the larger integer, and

2n² - 1 represents one less than twice the square of the small integer

we have

(n + 2)² = 2n² - 1        now solve for n

n² + 4n + 4 = 2n² - 1              (foil the left side)

0 = 2n² - n² - 4n - 4 - 1           (subtract everything on the left side from both

                                                sides)

0 = n² - 4n - 5                       (simplify)

0 =  (n - 5)(n + 1)                   (factor the equation)

So n = -1 and 5                      

***But -1 can't be an answer since problem said two consecutive POSITIVE odd numbers, so the only possible answer for n is 5.

5 is the smaller number, then 7 is the next odd integer

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844