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

1 Answer
Feb 6, 2016

Answer:

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.
#(x+1)^2+(x+2)=133#

You can see that we'll need to use FOIL to expand the first term.
#x^2+2x+1+(x+2)=133#

Then we combine terms.
#x^2+3x+3=133#

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

Factor.
#(x+13)(x-10)=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!