How do you solve $n^2+2n-24<=0$ ?

Question

1382 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-28T10:33:39

The answer is $n in [-6,4]$

Let's factorise the inequality

$n^2+2n-24=(n+6)(n-4)$

Let $f(n)=(n+6)(n-4)$

Now, we can build the sign chart

$color(white)(aaaa)$ $n$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-6$ $color(white)(aaaaa)$ $4$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $n+6$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $n-4$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(n)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(n)<=0$ when $n in [-6,4]$

How do you solve n^2+2n-24<=0n^2+2n-24<=0?