How do you solve $x^2-2x-8<0$ using a sign chart?

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Answer 1 · 2016-12-24T13:45:06

The answer is $x in ] -2,4 [$

We factorise the expression

$x^2-2x-8=(x+2)(x-4)$

Let, $f(x)=(x+2)(x-4)$

Now we can make the sign chart

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

$color(white)(aaaa)$ $x+2$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-4$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<0$ , when $x in ] -2,4 [$

How do you solve x^2-2x-8<0x^2-2x-8<0 using a sign chart?