SOLUTION: Find the smallest of three consecutive positive integers such that the product of the two smaller integers is 2 more than twice the largest integer.(an algebric solution can be acc

Question 196715: Find the smallest of three consecutive positive integers such that the product of the two smaller integers is 2 more than twice the largest integer.(an algebric solution can be accepted.)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find the smallest of three consecutive positive integers such that the product
of the two smaller integers is 2 more than twice the largest integer.
:
three consecutive positive integers:
x, x+1, x+2
:
"the product of the two smaller integers is 2 more than twice the largest integer.
:
x(x+1) = 2(x+2) + 2
x^2 + x = 2x + 4 + 2
x^2 + x = 2x + 6
Arrange as a quadratic equation
x^2 + x - 2x - 6 = 0
x^2 - x - 6 = 0
Factors to:
(x-3)(x+2) = 0
x = 3 is the positive solution
;
;
The numbers are 3, 4, 5
Check solution in the statement:
the product of the two smaller integers is 2 more than twice the largest integer.
3*4 = 2(5) + 2
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