Question

How do you solve `x^2-4x-32>0` using a sign chart?

Answer 1

The answer is `x in] -oo,-4 [ uu ] 8, oo[`

Explanation:

Let's factorise the expression `f(x)=x^2-4x-32=(x-8)(x+4)`

let's do the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaa)` `8` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `x-8` `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] -oo,-4 [ uu ] 8, oo[`

graph{x^2-4x-32 [-74, 74.05, -37, 37.14]}