# The difference between the squares of two consecutive integers is 11. Find the integers?

### 7 Answers

- ComoLv 78 months ago
:-

( x + 1 )² - x² = 11

x² + 2x + 1 - x² = 11

2x = 10

x = 5

Integers are 5 and 6

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- MICHAEL KLv 78 months ago
The answer is 5 and 6 where the square of 5 = 25 and the square of 6 = 36, so the remainder would be 11.

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- KrishnamurthyLv 78 months ago
The difference between the squares of two consecutive integers is 11.

(x + 1)^2 - x^2 = 11

(2x + 1) = 11

x = 5

The integers are 5 and 6.

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- PhilipLv 68 months ago
Let 2 consecutive integers be n, n+1. Then (n+1)^2 - n^2

= 11, ie., 2n+1 = 11, ie., n = 5 & consecutive integers are

5 and 6. 6^2 - 5^2 = 36 - 25 = 11.

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- sepiaLv 78 months ago
The difference between the squares of two consecutive integers is 11.

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

((n - 1) + n))((n - 1) - n)) = 11

(2n - 1)(-1) = 11

1 - 2n = 11

2n = -10

The two consecutive integers are -6 and -5.

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- Φ² = Φ+1Lv 78 months ago
The difference between n² and (n+1)² is |2n+1|.

The solutions for |2n+1|=11 are 5 and -6.

The consecutive integers are 5,6 and -6,-5.

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