Question

How do you determine three consecutive negative integers such that the square of third, added to the first, is 130?

Answer 1

The integers are `-14, -13 and -12.`

Explanation:

Let the consecutive integers be x, (x+1) and (x +2)
(Where x < 0)
`x + (x + 2)^2 = 130`

`x + x^2 + 4x + 4 = 130 " quadratic" rArr =0`

`x^2 + 5x -126 = 0`

The solutions have to be integers, so we will factorise rather than using completing the square or the formula.

Exploring factors of 126 gives:
`126 = 2xx3xx3xx7`

Find factors which differ by 5.
`2xx7 = 14 and 3xx3 =9 " 14 -9 =5"`

[The Factors are `(x + 14)(x -9)`]

`(x + 14)(x -9) = 0`
Solving gives: `x = -14 or x= 9`

We reject 9 because `x` has to a negative integer.

So, if `x = -14`, this is the first of the consecutive integers.

The integers are `-14, -13 and -12.`

Check: `-14 + (-12)^2 = -14 + 144 = 130!!!`