The sum of the squares of two consecutive negative odd integers is equal to 514. How do you find the two integers?

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Answer 1 · 2016-02-21T20:13:21

-15 and -17

Two odd negative numbers: $n$ $n$ and $n + 2$ $n+2$ .

The sum of squares=514:

$n^{2} + {(n + 2)}^{2} = 514$ $n^2+(n+2)^2=514$

$n^{2} + n^{2} + 4 n + 4 = 514$ $n^2+n^2+4n +4=514$

$2 n^{2} + 4 n - 510 = 0$ $2n^2+4n -510=0$

$n = \frac{- 4 \pm \sqrt{4^{2} - 4 \cdot 2 \cdot (- 510)}}{2 \cdot 2}$ $n=(-4+-sqrt(4^2-4*2*(-510)))/(2*2)$

$n = \frac{- 4 \pm \sqrt{16 + 4080}}{4}$ $n=(-4+-sqrt(16+4080))/4$

$n = \frac{- 4 \pm \sqrt{4096}}{4}$ $n=(-4+-sqrt(4096))/4$

$n = \frac{- 4 \pm 64}{4}$ $n=(-4+-64)/4$

$n = - \frac{68}{4} = - 17$ $n=-68/4=-17$ (because we want a negative number)

$n + 2 = - 15$ $n+2=-15$