Question

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

Answer 1

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

Explanation:

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 [`