How do you solve and write the following in interval notation: #x^2 - 2x - 8 > 0#?

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Answer 1 · 2017-11-02T10:03:55

The solution is #x in (-oo,-2) uu(4, +oo)#

Factorise the inequality

#x^2-2x-8=(x+2)(x-4)>0#

Let #f(x)=(x+2)(x-4)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)>0# when #x in (-oo,-2) uu(4, +oo)#

graph{(x+2)(x-4) [-31.62, 26.12, -9.97, 18.9]}