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

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Answer 1 · 2016-06-16T11:14:30

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

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