Question

Find 3 consecutive positive integers such that the product of the larger 2 is equal to twice the second integer plus twice the sum of all 3 integers

Answer 1

Step-by-step explanation:

x, x+1, x+2 are the 3 consecutive positive integers.

(x+1)(x+2) = 2(x+1) + 2×(x+x+1+x+2)

x² + 3x + 2 = 2x + 2 + 2×(3x + 3)

x² + 3x + 2 = 2x + 2 + 6x + 6 = 8x + 8

x² - 5x - 6 = 0

we can solve this by finding the factoring.

that means we find a, b so that

(x - a)(x - b) = x² - 5x - 6 = 0

because then x = a and x = b are the 2 solutions of the quadratic equation.

we have

x² - ax - bx + ab = x² - (a+b)x + ab = x² - 5x - 6

a+b = 5

ab = -6

that means, either a or b must be 6 and the other -1.

so we get

(x - 6)(x + 1) = 0

x = 6

or

x = -1

since we are only looking for positive integers, the x = -1 solution is invalid, and we get only x = 6 as solution.

therefore the 3 consecutive positive integers are

6, 7, 8