Answered

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14. Find three consecutive odd integers such that the product of the first and the third is 13 less
than ten times the second integer.

Sagot :

sqdancefan

Answer:

  {-1, 1, 3}  or  {7, 9, 11}

Step-by-step explanation:

Let x represent the middle integer. Then the first is (x-2) and the third is (x+2). The given relation is ...

  (x -2)(x +2) = 10x -13

  x^2 -4 = 10x -13 . . . . . . . eliminate parentheses

  x^2 -10x = -9 . . . . . . . . . . add 4-10x

  x^2 -10x +25 = 16 . . . . . add 25 to complete the square

  (x -5)^2 = 16 . . . . . . . . show the left side as a square

  x -5 = ±4 . . . . . . . . . . take the square root

  x = 5 ± 4 = {1, 9}

There are two solutions:

  {-1, 1, 3}  or  {7, 9, 11}

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