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]}

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