# How do you determine three consecutive positive integers such that the square of second, added to the third, is 133?

Feb 6, 2016

The integers are 10, 11, and 12. You set this up algebraically and use factoring to solve.

#### Explanation:

Let's set x as the first number in the sequence. We are told that the square of the second number (which is 1 more than x) added to the third number (which is 2 more than x) equals 133. The key is recognizing those parts in parentheses; once you recognize that the 2nd and 3rd numbers are just x+1 and x+2, it becomes a straight algebra problem.

So now let's set up our problem.
${\left(x + 1\right)}^{2} + \left(x + 2\right) = 133$

You can see that we'll need to use FOIL to expand the first term.
${x}^{2} + 2 x + 1 + \left(x + 2\right) = 133$

Then we combine terms.
${x}^{2} + 3 x + 3 = 133$

Move 133 over to set the equation equal to zero.
${x}^{2} + 3 x - 130 = 0$

Factor.
$\left(x + 13\right) \left(x - 10\right) = 0$

Set factors equal to zero and solve.
$x + 13 = 0 , x = - 13$
$x - 10 = 0 , x = 10$

Because they said that the integers are positive, we can ignore -13 and say that the first term is 10. You can check that the second integer (11) squared plus the third integer (12) equals 133, which it does!