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What is the smallest of 3 consecutive positvie integers if the product of the smaller two integers is 8 less than 4 times the largest integer?

User Bigmugcup
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Let x be the smallest of the three consecutive positive integers.
Then the other two integers are x + 1 and x + 2.

According to the problem, the product of the smaller two integers is 8 less than 4 times the largest integer. This can be written as:

x(x + 1) = 4(x + 2) - 8

Simplifying and solving for x, we get:

x^2 + x = 4x + 8 - 8
x^2 - 3x - 0 = 0

Using the quadratic formula, we get:

x = [3 ± sqrt(3^2 - 4(1)(0))] / 2
x = [3 ± sqrt(9)] / 2
x = (3 + 3) / 2 or x = (3 - 3) / 2
x = 3/2 or x = 0 (reject)

Since x must be a positive integer, the smallest of the three consecutive positive integers is x = 3/2, which is not an integer. Therefore, there is no solution to the problem as stated. Your welcome
User Jer In Chicago
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