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

1 Answer
Jun 16, 2016

Answer:

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!!!#