|
Question 228139: Which three consecutive integers have a product that is 800 times their sum?
Found 2 solutions by Alan3354, drj:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Which three consecutive integers have a product that is 800 times their sum?
x*(x+1)*(x+2) = 800*(3x+3)
x^3 + 3x^2 + 2x = 2400x + 2400
x^3 + 3x^2 - 2398x - 2400 = 0
x = -1
Divide by (x+1)
--> x^2 + 2x - 2400 = 0
(x + 50)*(x - 48) = 0
x = 48
x = -50
Answer by drj(1380)  (Show Source):
You can put this solution on YOUR website! Which three consecutive integers have a product that is 800 times their sum?
Step 1. Let n be the first integer.
Step 2. Let n+1 and n+2 be the next two consecutive integers.
Step 3. Let n+n+1+n+2=3n+3=3(n+1) be the sum of the three consecutive integers.
Step 4. Let n(n+1)(n+2) be the product of the three consecutive integers
Step 5. Then,n(n+1)(n+2)=800*3(n+1) since the product in Step 4 is 800 times their sum in Step 3.
Step 6. Solving n(n+1)(n+2)=800*3(n+1) yields the following steps:
Simplify by dividing n+1 to both sides of the equation
Subtract 2400 from both sides of the equation
Step 7. To solve, we can factor as follows:
 .
This implies that
 or 
and
 or 
We also could have use the quadratic formula given as
where a = 1, b=2, and c=-2400.
Step 9. With n=48, then n+1=49 and n+2=50. Check the relationship in Step 5 n(n+1)(n+2)=800(n+n+1+n+2) with these values.
 which is a true statement.
Step 10. With n=-50, then n+1=-49 and n+2=-48. Check the relationship in Step 5 n(n+1)(n+2)=800(n+n+1+n+2) with these values.
 which is a true statement.
Step 8. ANSWER: There are two sets of consecutive numbers. The first set is 48, 49, 50 and the second set is -48, -49, -50.
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
|
|
|
| |
