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

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Question 1108950: find three consecutive even integers such that the product of the first and third is 20 less than 10 times the second
Found 3 solutions by BumbleStar, addingup, TeachMath:
Answer by BumbleStar(7)   (Show Source): You can put this solution on YOUR website!

Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
n-2 * n+2 = 10(n)-20
n-2 * n+2 = 10n - 20
n(n+2) + (-2)(n+2) = 10n - 20
n^2 + 2n + (-2)(n + 2) = 10n - 20
n^2 + 2n + ((-2)n +(-2) * 2) = 10n - 20
n^2 + 2n + (-2)n - 4 = 10n - 20
n^2 - 4 = 10n - 20
n^2 - 4 - (10n - 20) = 0
n^2 - 4 + ((-1) * 10n + (-20)(-1)) = 0
n^2 - 4 + ((-10)n + (-20)(-1) = 0
n^2 - 4 + (-10)n + 20 = 0
n^2 + 16 + (-10)n = 0
n^2 + (-10)n + 16 = 0
n^2 + (-10)n = -16
The coefficient of n is -10. Divide by 2 and square (-10/2 = -5; -5^2 = 25), then add result to both sides of the equation:
n^2 + (-10)n + 25 = 9
now we have the equation in the form x^2 + bx + c, a perfect square we can factor:
(n-5)^2 = 9
Take the square root of both sides:
n-5 = 3
n = 8
your numbers are:
n - 2, n, n + 2
= 8 - 2, 8, 8 + 2
= 6, 8, 10 Three even consecutive integers
the product of the first and third is 20 less than 10 times the second:
6(10) = 10(8) - 20
60 = 80 - 20 Correct





Answer by TeachMath(96)   (Show Source): You can put this solution on YOUR website!
With the integers being N - 2, N, and N + 2, we get:

(N - 2)(N + 2) = 10N - 20

N^2 - 4 = 10N - 20	

N^2 - 10N + 16 = 0	

(N - 2)(N - 8) = 0	

N, or MIDDLE INTEGER = 2, or 8

Integers: 0, 2, and 4            OR             6, 8, and 10 

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