algebra 2 question

Question

find three consecutive odd integers such that the sum of the squares of the smallest integer and the largest interger is 129 more then the spuare of the median interger

how would you set this up and solve it

Answer 1

Let's try to solve this with a single unknown x. We know there are three consecutive odd integers, so the first one will be x, the second will be x + 2, and the last one will be x + 4.

Remember, this works because each odd integer is 2 away from the next odd integer, that's why we add 2.

The question says that the sum of the squares of the smallest integer and the largest integer are 129 more than the square of the median (middle) integer.

Let's translate this into equations one part at a time.

The square of the smallest integer is x² . The square of the largest integer is (x+4)². So then the sum of these will be x² + (x+4)². And this is 129 more than the square of the middle integer:

x² + (x+4)² = (x+2)² + 129

The reason we put the 129 on the right side is because the (x+2)² needs an extra 129 to be as large as x² + (x+4)².

Now we have our equation. Let's multiply everything out:

x² + (x² + 8x + 16) = (x² + 4x + 4) + 129

Remember the best way to solve a polynomial like this is to use factoring. So we group everything together and put it on the left side:

x² + 4x - 117 = 0

We have to figure out the factors of 117 to find out what our options are. We can immediately see that the digits of 117 add up to a multiple of 9, which means it is divisible by 9. 117/ 9 =13. Therefore 117 = 9*13. This seems to work for factoring.

Remember we need two numbers that multiply to -117 and add to 4, 13 and -9 work well.

(x + 13)(x - 9) = 0

x = -13 or x = 9

An integer can be negative so either one of these are valid answers for x. Let's choose x = 9. Then we also need to give the other odd consecutive integers. So your answer would end up being:

9, 11, and 13