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

answer: there are x+184 pages in olivia's book.

step-by-step explanation:

let the total number of pages in the book which olive has already read be x.

then if olive has 184 pages left to read, then the total number of pages are in olive's book= pages she has read+pages she has left to read

therefore, the total number of pages in olivia's book=

hence, there are x+184 pages in olivia's book.

translate k to n and reflect across the line containing jk.