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

1 Answer
Jan 28, 2017

Answer:

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

Explanation:

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]#