SOLUTION: Find three consecutive odd intergers such that the product of the first and the seconds exceeds the third by 8.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Find three consecutive odd intergers such that the product of the first and the seconds exceeds the third by 8.      Log On


   

Question 1083882: Find three consecutive odd intergers such that the product of the first and the seconds exceeds the third by 8.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Let x, %28x%2B2%29, and %28x%2B4%29 represent the three consecutive odd integers.
given:that the product of the first and the seconds exceeds the third by 8
x%28x%2B2%29=%28x%2B4%29%2B8
x%5E2%2B2x=x%2B12
x%5E2%2B2x-x-12=0
x%5E2%2Bx-12=0
x%5E2%2B4x-3x-12=0
%28x%5E2%2B4x%29-%283x%2B12%29=0
x%28x%2B4%29-3%28x%2B4%29=0
%28x+-+3%29%28x+%2B+4%29+=+0
solutions:
if %28x+-+3%29=+0->x=3
if %28x+%2B+4%29+=+0->x=-4-> disregard even
now find other two:
if first integer is 3, second is 3%2B2=5, and third one is 3%2B4=7
So the three consecutive odd integers are 3, 5, and 7.