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

Jun 16, 2016

The integers are $- 14 , - 13 \mathmr{and} - 12.$

#### Explanation:

Let the consecutive integers be x, (x+1) and (x +2)
(Where x < 0)
$x + {\left(x + 2\right)}^{2} = 130$

$x + {x}^{2} + 4 x + 4 = 130 \text{ quadratic} \Rightarrow = 0$

${x}^{2} + 5 x - 126 = 0$

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

Exploring factors of 126 gives:
$126 = 2 \times 3 \times 3 \times 7$

Find factors which differ by 5.
$2 \times 7 = 14 \mathmr{and} 3 \times 3 = 9 \text{ 14 -9 =5}$

[The Factors are $\left(x + 14\right) \left(x - 9\right)$]

$\left(x + 14\right) \left(x - 9\right) = 0$
Solving gives: $x = - 14 \mathmr{and} x = 9$

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 $- 14 , - 13 \mathmr{and} - 12.$

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