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| Question 221062:  what three consecutive numbers have a sum which is one fifth of their product?
 Answer by drj(1380)
      (Show Source): 
You can put this solution on YOUR website! What three consecutive numbers have a sum which is one fifth of their product? 
 Step 1.  Let n be one number
 
 Step 2.  Let n+1 and n+2 be the next two consecutive numbers.
 
 Step 3.  Then n+n+1+n+2=3n+3=3(n+1) be the sum of the three consecutive numbers.
 
 Step 4.  Then n*(n+1)*(n+2) be the product of the three consecutive numbers.
 
 Step 5.  Since the three consecutive numbers have a sum which is one fifth of their product, then
 
 
   
 Divide by n+1 to to sides of the equation
 
 
   
 Multiply by 5 to both sides of the equation to get rid of denominator
 
 
   
 
   
 Subtract 15 from both sides to get a quadratic
 
 
   
 Step 6.  Using the quadratic formula given as
 
 
   
 where a=1, b=1, and c=-15
 
 
 
 | Solved by pluggable solver: SOLVE quadratic equation with variable |  | Quadratic equation  (in our case  ) has the following solutons: 
 
  
 For these solutions to exist, the discriminant
  should not be a negative number. 
 First, we need to compute the discriminant
  :  . 
 Discriminant d=64 is greater than zero. That means that there are two solutions:
  . 
 
  
  
 Quadratic expression
  can be factored: 
  Again, the answer is: 3, -5.
Here's your graph:
 
  |  
 
 
  ,  ,   
 and
 
 
  ,  ,   
 Step 7.  ANSWER:  The consecutive numbers are 3, 4, 5  AND -5, -4, and -3
 
 I hope the above steps were helpful.
 
 For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
 
 Good luck in your studies!
 
 Respectfully,
 Dr J
 
 
 
 
 
 
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