SOLUTION: what three consecutive numbers have a sum which is one fifth of their product?

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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