PROBLEM
find three consecutive negative odd integers such that the product of the first and third is 71 more than 10 times the second.
SOLUTION
Let n be the first negative odd integer, let n - 2 be the second odd integer, and let n - 4 be the third odd integer.
Therefore, the product of the first and the third is (n)(n - 4) and 71 more than 10 times the second is 10(n - 2) + 71.
Writing the algebraic equation...
(n)(n - 4) = 10(n - 2) + 71
Using the distributive property...
n^2 - 4n = 10n - 20 + 71
Combining like terms...
n^2 - 4n - 10n + 20 - 71 = 0
Simplifying the equation...
n^2 - 14n - 51 = 0
Factoring the quadratic equation...
(n - 17)(n + 3) = 0
Solving for n...
n - 17 = 0, n + 3 = 0
n = 17, n = -3
Since n is negative, eliminate the n = 17 solution...
n = -3
n - 2 = -5
n - 4 = -7
Therefore, the three consecutive negative odd integers are -3, -5, and -7.
