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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)

User Gvasquez
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1 Answer

2 votes

Answer:

5 and 7

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

User Omatase
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