SOLUTION: Find three consecutive even integers such that the product of the first and second is 20 less than 10 times the third

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Question 1200102: Find three consecutive even integers such that the product of the first and second is 20 less than 10 times the third
Found 2 solutions by math_tutor2020, josgarithmetic:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Consecutive integers follow one after another.
Examples:
6,7,8
22,23,24

Consecutive even integers are when all of the values are even, and they are as close as possible.
Examples:
2,4,6
100,102,104

Let x be some integer. It might be even or odd.
To guarantee it is always even, we double it to 2x.
Eg:
x = 5 doubles to 2x = 2*5 = 10
x = 6 doubles to 2x = 2*6 = 12

2x = first even integer
2x+2 = even integer right after 2x
2x+2+2 = 2x+4 = even integer right after 2x+2

In short,
first = 2x
second = 2x+2
third = 2x+4

The sequence {2x,2x+2,2x+4} represents a set of three consecutive even integers, where x is any integer.

We then have the following information " the product of the first and second is 20 less than 10 times the third".

Let's translate that into an equation and solve for x.

first*second = 10*third - 20
2x*(2x+2) = 10*(2x+4) - 20
4x^2+4x = 20x+40 - 20
4x^2+4x = 20x+20
4x^2+4x-20x-20 = 0
4x^2-16x-20 = 0
4(x^2-4x-5) = 0
x^2-4x-5 = 0
(x-5)(x+1) = 0
x-5 = 0 or x+1 = 0
x = 5 or x = -1

Optionally you could use the quadratic formula to solve for x.
Let me know if you need to see these steps.

If x = 5, then,
  • 2x = 2*5 = 10 is the first even integer
  • 2x+2 = 2*5+2 = 10+2 = 12 is the second even integer
  • 2x+4 = 2*5+4 = 10+4 = 14 is the third even integer
We end up with the set {10,12,14} as one solution.

If x = -1, then,
  • 2x = 2*(-1) = -2 is the first even integer
  • 2x+2 = 2*(-1)+2 = -2+2 = 0 is the second even integer
  • 2x+4 = 2*(-1)+4 = -2+4 = 2 is the third even integer
We end up with the set {-2,0,2} as the other solution.


Check for the first solution:
first*second = 10*third - 20
10*12 = 10*14 - 20
120 = 140 - 20
120 = 120
The first solution set is confirmed.

Check for the second solution:
first*second = 10*third - 20
-2*0 = 10*2 - 20
0 = 20 - 20
0 = 0
The second solution set is confirmed.

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
n, n+2, n+4
n%28n%2B2%29=10%28n%2B4%29-20
-
n%5E2%2B2n=10n%2B40-20
n%5E2-8n=20
n%28n-8%29=20
Try to avoid quadratic formula general solution method.

Two factors differing by 8 to give product 20:
10 and 2.
n=10.

The consecutive even integers: 10, 12, 14.