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Consider three consecutive positive integers, such that the sum of the
squares of the two larger integers is 5 more than 40 times the smaller
one. Find the smaller integer.

Respuesta :

sqdancefan

Answer:

  17

Step-by-step explanation:

Let x represent the smaller integer. Then we have ...

  (x +1)² +(x +2)² = 40x +5

  2x² +6x +5 = 40x +5

  x² -17x = 0 . . . . . subtract (40x+5), divide by 2

  x(x -17) = 0 . . . . . factor

The solution of interest is x = 17.

The smaller integer is 17.

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