What are three consecutive integers whose product is 480 more than their sum?
1 answer:
Answer:
The three numbers are 7 8 and 9
Step-by-step explanation:
Givens
- Let the first number be n - 1
- Let the second number be n
- Let the third number = n + 1
Equation
(n - 1)(n)(n + 1) - (n-1 + n + n+1) = 480
Solution
Multiply (n - 1) and (n + 1) = (n - 1)*(n + 1) = n^2 - 1
Multiply the second integer by the result of the first and third: n (n^2 - 1)
Add the three integers together: (x - 1) + (n - 1) + n = 3n Combine these 2 steps
n(n^2 - 1) - 3n = 480 Remove the brackets
n^3 - n - 3n = 480
n^3 - 4n = 480
n^3 - 4n - 480 = 0
Graph
The graph shows that the intercept point is n =8. This is the only way I can see to solve this cubic. There are no other real roots.
Answer
n - 1 = 7
n = 8
n + 1 = 9
Check
Product 7*8*9 = 504
Sum = 7 + 8 + 9 = 24
504 - 24 = 480 Which checks.
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