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Dennis_Churaev [7]
3 years ago
6

Find three consecutive positive integers such that the product of the first and the third is 11 more than the second

Mathematics
1 answer:
Len [333]3 years ago
8 0

Answer:

3, 4, 5

Step-by-step explanation:

If the integers are x, x+1, and x+2, then:

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

x² + 2x = x + 12

x² + x − 12 = 0

(x + 4) (x − 3) = 0

x = -4 or 3

Since x is positive, x = 3.  So the three integers are 3, 4, and 5.

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