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3 years ago

Consider two consecutive positive integers such that the square of the second integer added to 3 times the first is equal to 37.

Mathematics

1 answer:

wlad13 [49]3 years ago

4 0

Am I supposed to consider or solve? Let's solve.

Let's call our integers x and x+1

square of the second integer added to 3 times the first is 37

(x+1)² + 3x = 37

x² + 2x + 1 + 3x - 37 = 0

x² + 5x - 36 = 0

(x+9)(x-4) = 0

Answer: x = -9 or x=4

Check:

x=-9

(-9+1)² + 3(-9) = 64 - 27 = 37 good

x=4

(4+1)² + 3(4) = 25 + 12 = 37 good

Two solutions.

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