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:

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

$x + 1 = 0$

$x + 1 - 1 = 0 - 1$

$x = -1$