How do you find two consecutive odd integers whose product is 1 less than 6 times their sum?

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Answer 1 · 2016-11-26T19:41:26

There are two sets of consecutive odd integers which meet the requirements of this problem:

11 and 13

or

-1 and 1

First, let's define our numbers in mathematical terms.

Let #x# be the first odd integer. Then, because we are looking for consecutive odd integers, we can let the second integer be #x+2#.

So this problem can now be rewritten as:

#x*(x + 2) = (6*(x + (x+2))) - 1#

#x^2 + 2x = (6*(2x + 2)) - 1#

#x^2 + 2x = 12x + 12 - 1#

#x^2 + 2x = 12x + 11#

Next create a quadratic equation:

#x^2 + 2x - 12x - 11 = 12x + 11 - 12x - 11#

#x^2 - 10x - 11 = 0#

Factoring this gives:

#(x - 11)(x + 1) = 0#

Solving first for #(x - 11)# gives:

#((x - 11)(x + 1))/(x + 1) = 0/(x + 1)#

#x - 11 = 0#

#x - 11 + 11 = 0 + 11#

#x = 11#

Solving for (x + 1) gives:

#((x - 11)(x + 1))/(x - 11) = 0/(x -11)#

#x + 1 = 0#

#x + 1 - 1 = 0 - 1#

#x = -1#